Difference between revisions of "Table of indefinite integrals" - Rhea

Latest revision as of 17:06, 26 February 2015

Indefinite Integrals

General Rules
$\int a d x = a x$
$\int a f ( x ) d x = a \int f ( x ) d x$
$\int ( u \pm v \pm w \pm \cdot \cdot \cdot ) d x = \int u d x \pm \int v d x \pm \int w d x \pm \cdot \cdot \cdot$
$\int u d v = u v - \int v d u$
$\int f ( a x ) d x = \frac{1}{a} \int f ( u ) d u$
$\int F \{ f ( x ) \} d x = \int F ( u ) \frac{dx}{du} d u = \int \frac{F ( u )}{f^{'} ( x )} d u \qquad u = f ( x )$
$\int u^n d u = \frac{u^{n+1}}{n+1} \qquad n \neq -1$
$\int \frac{d u}{u} = \ln u \ ( if \ u > 0 ) \ \text{or} \ln {-u} \ ( \text{if} \ u < 0 ) = \ln \left \| u \right \|$
$\int e^u d u = e^u$
Click here for more general rules.
Transformations of the independent variable
$\int F( a x + b) d x =\frac{1}{a} \int F( u) d u \qquad u = a x + b$
$\int F( \sqrt {a x + b} ) d x =\frac{2}{a} \int u F( u) d u \qquad u = \sqrt {a x + b}$
$\int F( \sqrt [n] {a x + b} ) d x = \frac{n}{a} \int u^{n-1} F( u) d u \qquad u = \sqrt [n] {a x + b}$
$\int F( \sqrt {a^2 - x^2} ) d x =a \ \int F( a \cos u) \ \cos u \ d u \qquad x = a \sin u$
$\int F( \sqrt {x^2 + a^2} ) d x =a \ \int F \left ( \frac {a}{\cos u} \right ) \frac {1}{\cos ^2 u} \ d u \qquad x = a \tan u$
$\int F( \sqrt {x^2 - a^2} ) d x =a \ \int F \left ( a \tan u \right ) \frac {\tan u}{\cos u} \ d u \qquad x = \frac {a}{\cos u}$
$\int F( e ^{a x}) d x = \frac {1}{a} \int \frac {F(u)}{u} \ d u \qquad u = e^{a x}$
$\int F( \ln x ) d x = \int F(u)\ e^u \ d u \qquad u = \ln x$
$\int F\left ( \arcsin \frac{x}{a} \right) d x = a \int F(u)\ \cos u \ d u \qquad u = \arcsin \frac {x}{a}$
$\int F\left ( \sin x ,\cos x \right) d x = 2 \int F \left( \frac {2 u}{1 + u^2}, \frac {1 - u^2}{1+u^2} \right)\ \frac {d u}{1+ u^2} \qquad u = \tan \frac {x}{2}$
Integrals with ax +b
$\int \frac {d x}{ ax + b} = \frac {1}{a} \ln (ax +b)$
$\int \frac {x d x}{ ax + b} = \frac {x}{a} - \frac{b}{a^2} \ln (ax +b)$
$\int \frac {x^2 d x}{ ax + b} = \frac {(ax+b)^2}{2a^3} - \frac {2b(ax+b) }{a^3} + \frac{b^2}{a^3} \ln (ax +b)$
$\int \frac {d x}{\sqrt{a x +b}} = \frac {2\sqrt{ax+b}}{a}$
$\int \frac {x d x}{\sqrt{a x + b}} = \frac {2(ax-2b)}{3a^2}\sqrt{ax+b}$
$\int \frac {x^2 d x}{\sqrt{a x + b}} = \frac {2(3a^2x^2-4abx + 8b^2)}{15a^3}\sqrt{ax+b}$
Click here for more integrals with ax+b.

More

Back to Collective Table of Formulas

@@ Line 15: / Line 15: @@
 {|
 |-
-! style="background-color: rgb(228, 188, 126); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Table of Indefinite Integrals
+! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | [[Table_of_indefinite_integrals_general_rules|General Rules]]
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 1 General Rules
 |-
 |<math> \int a d x =  a x </math>
@@ Line 39: / Line 37: @@
 |<font size= 4> Click [[Table_of_indefinite_integrals_general_rules|here]] for [[Table_of_indefinite_integrals_general_rules|more general rules]]. </font size>
 |-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 2 Important Transformations
+! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Transformations of the independent variable
 |-
 |<math> \int F( a x + b) d x =\frac{1}{a} \int F( u) d u  \qquad  u = a x + b</math>
@@ Line 62: / Line 60: @@
 |-
 |-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Integrals with ax +b
+! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | [[Table_of_indefinite_integrals_axplusb|Integrals with ax +b]]
 |-
 |<math> \int \frac {d x}{ ax + b} = \frac {1}{a} \ln (ax +b)</math>
@@ Line 82: / Line 80: @@
 * [[Table_of_indefinite_integrals_axplusb_and_pxplusq|Integrals with ax+b and px+q]].
 * [[Table_of_indefinite_integrals_xnplusan|Integrals with <math>x^n+a^n</math>]]
-* [[Table_of_indefinite_integrals_xnminusan|Integrals with <math>x^n-a^n</math>]] </font size>
+* [[Table_of_indefinite_integrals_xnminusan|Integrals with <math>x^n-a^n</math>]]
 * [[Table_of_indefinite_integrals_cosine_sine|Integrals with <math>\cos x</math> and/or <math>\sin x</math>]]
 * [[Table_of_indefinite_integrals_cosine_sine|Integrals with <math>\cos x</math> and/or <math>\sin x</math>]]
@@ Line 93: / Line 91: @@
 *[[Table_of_indefinite_integrals_ch|Integrals with hyperbolic cosine (ch x)]]
 *[[Table_of_indefinite_integrals_th|Integrals with hyperbolic tangent (th x)]]
-*[[Table_of_indefinite_integrals_coth|Integrals with hyperbolic cotangent (both x)]]
+*[[Table_of_indefinite_integrals_coth|Integrals with hyperbolic cotangent (coth x)]]
-{|
+*[[Table_of_indefinite_integrals_x_inverse|Integrals with 1/x]]
+*[[Table_of_indefinite_quadratic|Integrals with <math>ax^2+bx+c</math>]]
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 32 Integrals of Hyperbolic Inverse functions : arg sh ax
-|-
-|-
-| <math> \int\arg sh\dfrac{x}{a}dx=x\arg sh\dfrac{x}{a}-\sqrt{x^{2}+a^{2}} </math>
-|-
-| <math> \int x\arg sh\dfrac{x}{a} dx=\biggl(\dfrac{x^{2}}{2}+\dfrac{a^{2}}{4}\biggl)\arg sh\dfrac{x}{a}-\dfrac{x\sqrt{x^{2}+a^{2}}}{4} </math>
-|-
-| <math> \int x^{2}\arg sh\dfrac{x}{a} dx=\dfrac{x^{3}}{3}\arg sh\dfrac{x}{a}+\dfrac{(2a^{2-}x^{2})\sqrt{x^{2}+a^{2}}}{9} </math>
-|-
-| <math> \int\dfrac{\arg sh\dfrac{x}{a}}{x}dx=\Biggl\{\begin{array}{c}
-\dfrac{x}{a}-\dfrac{(\dfrac{x}{a})^{3}}{2\cdot3\cdot3}+\dfrac{1\cdot3(\dfrac{x}{a})^{5}}{2\cdot4\cdot5\cdot5}-\dfrac{1\cdot3\cdot5(\dfrac{x}{a})^{7}}{2\cdot4\cdot6\cdot7\cdot7}+\cdots,|x|<a\\
-\dfrac{\ln^{2}(\dfrac{2x}{a})}{2}-\dfrac{(\dfrac{a}{x})^{2}}{2\cdot2\cdot2}+\dfrac{1\cdot3(\dfrac{a}{x})^{4}}{2\cdot4\cdot4\cdot4}-\dfrac{1\cdot3\cdot5(\dfrac{a}{x})^{6}}{2\cdot4\cdot6\cdot6\cdot6}+\cdots, x>a\\
-\dfrac{\ln^{2}(\dfrac{-2x}{a})}{2}+\dfrac{(\dfrac{a}{x})^{2}}{2\cdot2\cdot2}-\dfrac{1\cdot3(\dfrac{a}{x})^{4}}{2\cdot4\cdot4\cdot4}+\dfrac{1\cdot3\cdot5(\dfrac{a}{x})^{6}}{2\cdot4\cdot6\cdot6\cdot6}+\cdots, x<-a\end{array} </math>
-|-
-| <math> \int\dfrac{\arg sh\dfrac{x}{a}}{x^{2}}dx=-\dfrac{\arg sh\dfrac{x}{a}}{x}-\dfrac{1}{a}\ln\Biggl(\dfrac{a+\sqrt{x^{2}+a^{2}}}{x}\Biggl) </math>
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 33 Integrals of Hyperbolic Inverse functions : arg ch ax
-|-
-|-
-| <math> \int\arg ch\dfrac{x}{a}dx=\begin{cases}
-\dfrac{x\arg ch\dfrac{x}{a}-\sqrt{x^{2}-a^{2}}}{x\arg sh\dfrac{x}{a}+\sqrt{x^{2}-a^{2}}} & .\end{cases} </math>
-|-
-| <math> \int x\arg ch\dfrac{x}{a} dx=\begin{cases}
-\dfrac{\frac{1}{4}(2x^{2}-a^{2})\arg ch\dfrac{x}{a}-\frac{1}{4}x\sqrt{x^{2}-a^{2}}}{\frac{1}{4}(2x^{2}-a^{2})\arg ch\dfrac{x}{a}+\frac{1}{4}x\sqrt{x^{2}-a^{2}}} & .\end{cases} </math>
-|-
-| <math> \int x^{2}\arg ch\dfrac{x}{a} dx=\begin{cases}
-\dfrac{\frac{1}{3}x^{3}\arg ch\dfrac{x}{a}-\frac{1}{9}(x^{2}+2a^{2})\sqrt{x^{2}-a^{2}}}{\frac{1}{3}x^{3}\arg ch\dfrac{x}{a}+\frac{1}{9}(x^{2}+2a^{2})\sqrt{x^{2}-a^{2}}} & .\end{cases} </math>
-|-
-| <math> \int\dfrac{\arg ch\dfrac{x}{a}}{x}dx=\pm\Biggl[\dfrac{\ln^{2}(\dfrac{2x}{a})}{2}+\dfrac{(\dfrac{a}{x})^{2}}{2\cdot2\cdot2}+\dfrac{1\cdot3(\dfrac{a}{x})^{4}}{2\cdot4\cdot4\cdot4}+\dfrac{1\cdot3\cdot5(\dfrac{a}{x})^{6}}{2\cdot4\cdot6\cdot6\cdot6}+\cdots,\Biggl] </math>
-|-
-| <math> \int\dfrac{\arg ch\dfrac{x}{a}}{x^{2}}dx=-\dfrac{\arg ch\dfrac{x}{a}}{x}\mp\dfrac{1}{a}\ln\Biggl(\dfrac{a+\sqrt{x^{2}+a^{2}}}{x}\Biggl) </math>
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 34 Integrals of Hyperbolic Inverse functions : arg th ax
-|-
-|-
-| <math> \int\arg th\dfrac{x}{a}dx=x\arg th\dfrac{x}{a}+\dfrac{a}{2}\ln(a^{2}-x^{2}) </math>
-|-
-| <math> \int x\arg th\dfrac{x}{a} dx=\dfrac{ax}{2}+\frac{1}{2}(x^{2}-a^{2})\arg th\dfrac{x}{a} </math>
-|-
-| <math> \int x^{2}\arg th\dfrac{x}{a} dx=\dfrac{ax^{2}}{6}+\frac{a^{3}}{6}\ln(a^{2}-x^{2})+\dfrac{x^{3}}{3}\arg th\dfrac{x}{a} </math>
-|-
-| <math> \int\dfrac{\arg th\dfrac{x}{a}}{x}dx=\dfrac{x}{a}+\dfrac{(\dfrac{x}{a})^{3}}{3^{2}}+\dfrac{(\dfrac{x}{a})^{5}}{5^{2}}+\cdots </math>
-|-
-| <math> \int\dfrac{\arg th\dfrac{x}{a}}{x^{2}}dx=-\dfrac{\arg th\dfrac{x}{a}}{x}+\dfrac{1}{2a}\ln\Biggl(\dfrac{x^{2}}{a^{2}-x^{2}}\Biggl) </math>
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 35 Integrals of Hyperbolic Inverse functions : arg coth ax
-|-
-|-
-| <math> \int\arg coth\dfrac{x}{a}dx=x\arg coth x+\dfrac{a}{2}\ln(x^{2}-a^{2}) </math>
-|-
-| <math> \int x\arg coth\dfrac{x}{a} dx=\dfrac{ax}{2}+\frac{1}{2}(x^{2}-a^{2})\arg coth\dfrac{x}{a} </math>
-|-
-| <math> \int x^{2}\arg coth\dfrac{x}{a} dx=\dfrac{ax^{2}}{6}+\frac{a^{3}}{6}\ln(x^{2}-a^{2})+\dfrac{x^{3}}{3}\arg coth\dfrac{x}{a} </math>
-|-
-| <math> \int\dfrac{\arg coth\dfrac{x}{a}}{x}dx=-\Biggl(\dfrac{a}{x}+\dfrac{(\dfrac{a}{x})^{3}}{3^{2}}+\dfrac{(\dfrac{a}{x})^{5}}{5^{2}}+\cdots\Biggl) </math>
-|-
-| <math> \int\dfrac{\arg coth\dfrac{x}{a}}{x^{2}}dx=-\dfrac{\arg coth\dfrac{x}{a}}{x}+\dfrac{1}{2a}\ln\Biggl(\dfrac{x^{2}}{x^{2}-a^{2}}\Biggl) </math>
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | 36 Integrals of a/x
-|-
-| <math> \int\arg ch\dfrac{a}{x}dx=\begin{cases}
-\dfrac{x\arg ch\dfrac{a}{x}+\arcsin\dfrac{x}{a}}{x\arg ch\dfrac{a}{x}-\arcsin\dfrac{x}{a}} & .\end{cases} </math>
-|-
-| <math> \int x\arg ch\dfrac{a}{x} dx=\begin{cases}
-\dfrac{\frac{1}{2}x^{2}\arg ch\dfrac{a}{x}-\dfrac{1}{2}a\sqrt{a^{2}-x^{2}}}{\frac{1}{2}x^{2}\arg ch\dfrac{a}{x}+\dfrac{1}{2}a\sqrt{a^{2}-x^{2}}} & .\end{cases} </math>
-|-
-| <math> \int\dfrac{\arg ch\dfrac{a}{x}}{x}dx=\begin{cases}
-\dfrac{\dfrac{-\frac{1}{2}\ln(\dfrac{a}{x})\ln(\dfrac{4a}{x})}{2}-\dfrac{(\dfrac{x}{a})^{2}}{2\cdot2\cdot2}-\dfrac{1\cdot3(\dfrac{x}{a})^{4}}{2\cdot4\cdot4\cdot4}-\cdots}{\dfrac{\frac{1}{2}\ln(\dfrac{a}{x})\ln(\dfrac{4a}{x})}{2}+\dfrac{(\dfrac{x}{a})^{2}}{2\cdot2\cdot2}+\dfrac{1\cdot3(\dfrac{x}{a})^{4}}{2\cdot4\cdot4\cdot4}+\cdots} & .\end{cases} </math>
-|-
-| <math> \int\arg sh\dfrac{a}{x}dx=x\arg sh\dfrac{a}{x}\pm\arg sh\dfrac{x}{a} </math>
-|-
-| <math> \int x\arg sh\dfrac{a}{x} dx=\dfrac{x^{2}}{2}\arg sh\dfrac{a}{x}\pm\dfrac{1}{2}a\sqrt{a^{2}+x^{2}} </math>
-|-
-| <math> \int\dfrac{\arg sh\dfrac{a}{x}}{x}dx=\Biggl\{\begin{array}{c}
-\dfrac{\frac{1}{2}\ln(\dfrac{x}{a})\ln(\dfrac{4a}{x})}{2}+\dfrac{(\dfrac{x}{a})^{2}}{2\cdot2\cdot2}-\dfrac{1\cdot3(\dfrac{x}{a})^{4}}{2\cdot4\cdot4\cdot4}+\cdots\\
-\dfrac{\frac{1}{2}\ln(\dfrac{x}{a})\ln(\dfrac{4a}{x})}{2}-\dfrac{(\dfrac{x}{a})^{2}}{2\cdot2\cdot2}+\dfrac{1\cdot3(\dfrac{x}{a})^{4}}{2\cdot4\cdot4\cdot4}+\cdots\\
--\frac{a}{x}+\dfrac{(\dfrac{a}{x})^{3}}{2\cdot3\cdot3}-\dfrac{1\cdot3(\dfrac{a}{x})^{5}}{2\cdot4\cdot5\cdot5}+\cdots\end{array} </math>
-|-
-| <math> \int x^{m}\arg sh\dfrac{x}{a} dx=\dfrac{x^{m+1}}{m+1}\arg sh\dfrac{x}{a}-\dfrac{1}{m+1} \int\dfrac{x^{m+1}}{\sqrt{x^{2}+a^{2}}}dx </math>
-|-
-| <math> \int x^{m}\arg ch\dfrac{x}{a} dx=\begin{cases}
-\dfrac{\dfrac{x^{m+1}}{m+1}\arg ch\dfrac{x}{a}-\dfrac{1}{m+1} \int\dfrac{x^{m+1}}{\sqrt{x^{2}-a^{2}}}dx}{\dfrac{x^{m+1}}{m+1}\arg ch\dfrac{x}{a}+\dfrac{1}{m+1} \int\dfrac{x^{m+1}}{\sqrt{x^{2}-a^{2}}}dx} & .\end{cases} </math>
-|-
-| <math> \int x^{m}\arg th\dfrac{x}{a} dx=\dfrac{x^{m+1}}{m+1}\arg th\dfrac{x}{a}-\dfrac{a}{m+1} \int\dfrac{x^{m+1}}{a^{2}-x^{2}}dx </math>
-|-
-| <math> \int x^{m}\arg coth\dfrac{x}{a} dx=\dfrac{x^{m+1}}{m+1}\arg coth\dfrac{x}{a}-\dfrac{a}{m+1} \int\dfrac{x^{m+1}}{a^{2}-x^{2}}dx </math>
-|-
-| <math> \int x^{m}\arg ch\dfrac{a}{x} dx=\begin{cases}
-\dfrac{\dfrac{x^{m+1}}{m+1}\arg ch\dfrac{a}{x}+\dfrac{a}{m+1} \int\dfrac{x^{m}}{\sqrt{a^{2}-x^{2}}}dx}{\dfrac{x^{m+1}}{m+1}\arg ch\dfrac{a}{x}-\dfrac{a}{m+1} \int\dfrac{x^{m}}{\sqrt{a^{2}-x^{2}}}dx} & .\end{cases} </math>
-|-
-| <math> \int x^{m}\arg sh\dfrac{a}{x} dx=\dfrac{x^{m+1}}{m+1}\arg sh\dfrac{x}{a}\pm\dfrac{a}{m+1} \int\dfrac{x^{m}}{\sqrt{x^{2}+a^{2}}}dx </math>
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math> x^{2}-a^{2},x^{2}<a^{2} </math>
-|-
-| <math> \int \dfrac{dx}{a^{2}-x^{2}} = \dfrac{1}{2a}\ln\left(\dfrac{a+x}{a-x}\right) \qquad o\grave{u}\qquad \dfrac{1}{a} Arg tg \dfrac{x}{a}</math>
-|-
-| <math> \int \dfrac{xdx}{a^2 - x^2} = -\dfrac{1}{2} \ln\left({a^2-x^2}\right) </math>
-|-
-| <math> \int \dfrac{x^2 dx}{a^2 - x^2} = -x + \dfrac{a}{2}\ln\left(\dfrac{a+x}{a-x}\right)</math>
-|-
-| <math> \int \dfrac{x^3 dx}{a^2 - x^2} = -\dfrac{x^2}{2} - \dfrac{a^2}{2}\ln\left({a^2-x^2}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x\left(a^2-x^2\right)} = \dfrac{1}{2a^2}\ln\left(\dfrac{x^2}{a^2-x^2}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(a^2-x^2\right)} =  -\dfrac{1}{a^2x} + \dfrac{1}{2a^3}\ln\left(\dfrac{a+x}{a-x}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x^3\left(a^2-x^2\right)} = -\dfrac{1}{2a^2x^2} + \dfrac{1}{2a^4}\ln\left(\dfrac{x^2}{a^2-x^2}\right)</math>
-|-
-| <math> \int \dfrac{dx}{\left(a^2-x^2\right)^2} = \dfrac{x}{2a^2\left(a^2-x^2\right)} + \dfrac{1}{4a^3}\ln\left(\dfrac{a+x}{a-x}\right)</math>
-|-
-| <math> \int \dfrac{xdx}{\left(a^2-x^2\right)^2} = \dfrac{1}{2\left(a^2-x^2\right)}</math>
-|-
-| <math> \int \dfrac{x^2dx}{\left(a^2-x^2\right)^2} = \dfrac{x}{2\left(a^2-x^2\right)} - \dfrac{1}{4a}\ln\left(\dfrac{a+x}{a-x}\right)</math>
-|-
-| <math> \int \dfrac{x^3dx}{\left(a^2-x^2\right)^2} = \dfrac{a^2}{2\left(a^2-x^2\right)} - \dfrac{1}{2}\ln\left(a^2-x^2\right)</math>
-|-
-| <math> \int \dfrac{dx}{x\left(a^2-x^2\right)^2} = \dfrac{1}{2a^2\left(a^2-x^2\right)} - \dfrac{1}{2a^4}\ln\left(\dfrac{x^2}{a^2-x^2}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(a^2-x^2\right)^2} = \dfrac{-1}{a^4x} + \dfrac{x}{2a^4\left(a^2-x^2\right)} - \dfrac{3}{4a^5}\ln\left(\dfrac{a+x}{a-x}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x^3\left(a^2-x^2\right)^2} = \dfrac{-1}{2a^4x^2} + \dfrac{1}{2a^4\left(a^2-x^2\right)} + \dfrac{1}{a^6}\ln\left(\dfrac{x^2}{a^2-x^2}\right)</math>
-|-
-| <math> \int \dfrac{dx}{\left(a^2-x^2\right)^n} = \dfrac{x}{2\left(n-1\right)a^2\left(a^2-x^2\right)^{n-1}} + \dfrac{2n-3}{\left(2n-2\right)a^2} \int\dfrac{dx}{\left(a^2-x^2\right)^{n-1}}</math>
-|-
-| <math> \int \dfrac{xdx}{\left(a^2-x^2\right)^n} = \dfrac{1}{2\left(n-1\right)\left(a^2-x^2\right)^{n-1}}</math>
-|-
-| <math> \int \dfrac{dx}{x\left(a^2-x^2\right)^n} = \dfrac{1}{2\left(n-1\right)\left(a^2-x^2\right)^{n-1}} + \dfrac{1}{a^2}\int\dfrac{dx}{x\left(a^2-x^2\right)^{n-1}} </math>
-|-
-| <math> \int \dfrac{x^mdx}{\left(a^2-x^2\right)^n} = a^2\int\dfrac{x^{m-2}dx}{\left(a^2-x^2\right)^{n-1}} - \int\dfrac{x^{m-2}dx}{\left(a^2-x^2\right)^{n-1}}</math>
-|-
-| <math> \int \dfrac{dx}{x^m\left(a^2-x^2\right)^n} = \dfrac{1}{a^2}\int\dfrac{dx}{x^m\left(a^2-x^2\right)^{n-1}} + \dfrac{1}{a^2}\int\dfrac{dx}{x^{m-2}\left(a^2-x^2\right)^n} </math>
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math> ax^2 + bx + c </math>
-|-
-| <math> \int \dfrac{dx}{ax^2 + bx + c} =
-\begin{cases}
-\dfrac{2}{\sqrt{4ac-b^2}} \arctan \dfrac{2ax+b}{\sqrt{4ac-b^2}} \\
-\dfrac{1}{\sqrt{b^2-4ac}} \ln\left(\dfrac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right)
-\end{cases} </math>
-|-
-| <math> Si\quad b^2 = 4ac, ax^2 + bx + c = a\left(x+ b/2a\right)^2 et\quad on\quad peut\quad utiliser\quad les\quad r\acute{e}sultats\quad des\quad pages \quad60-61. </math>
-|-
-| <math> \quad Si \quad b = 0 \quad utiliser \quad les \quad r\acute{e}sultats \quad de \quad la \quad page \quad 64. \quad Si \quad a \quad ou \quad c = 0,\quad utiliser \quad les </math>
-|-
-| <math> \quad r\acute{e}sultats \quad des \quad pages \quad 60-61. </math>
-|-
-| <math> \int \dfrac{xdx}{ax^2 + bx + c} = \dfrac{1}{2a} \ln\left(ax^2+bx+c\right) - \dfrac{b}{2a}\int \dfrac{dx}{ax^2 + bx + c} </math>
-|-
-| <math> \int \dfrac{x^2dx}{ax^2 + bx + c} = \dfrac{x}{a} - \dfrac{b}{2a^2}\ln\left(ax^2+bx+c\right) + \dfrac{b^2-2ac}{2a^2} \dfrac{b}{2a}\int \dfrac{dx}{ax^2 + bx + c} </math>
-|-
-| <math> \int \dfrac{x^{m}dx}{ax^2 + bx + c} = \dfrac{x^{m-1}}{\left(m-1\right)a} - \dfrac{c}{a}\int \dfrac{x^{m-2}dx}{ax^2 + bx + c} - \dfrac{b}{a}\int \dfrac{x{m-1}dx}{ax^2 + bx + c} </math>
-|-
-| <math> \int \dfrac{dx}{x\left(ax^2 + bx + c\right)} = \dfrac{1}{2c} \ln\left(\dfrac{x^2}{ax^2+bx+c}\right) - \dfrac{b}{2c} \int \dfrac{dx}{ax^2+bx+c} </math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(ax^2 + bx + c\right)} = \dfrac{b}{2c^2} \ln\left(\dfrac{ax^2+bx+c}{x^2}\right) - \dfrac{1}{cx} + \dfrac{b^2-2ac}{2c^2} \int \dfrac{dx}{ax^2+bx+c} </math>
-|-
-| <math> \int \dfrac{dx}{x^n\left(ax^2 + bx + c\right)} = - \dfrac{1}{\left(n-1\right)cx^{n-1}} - \dfrac{b}{c}\int \dfrac{dx}{x^{n-1}\left(ax^2+bx+c\right)} - \dfrac{a}{c}\int \dfrac{dx}{x^{n-2}\left(ax^2+bx+c\right)} </math>
-|-
-| <math> \int \dfrac{dx}{\left(ax^2+bx+c \right)^2} = \dfrac{2ax+b}{\left(4ac-b^2 \right) \left(ax^2+bx+c \right)} + \dfrac{2a}{4ac-b^2} \int \dfrac{dx}{ax^2+bx+c} </math>
-|-
-| <math> \int \dfrac{xdx}{\left(ax^2+bx+c^2 \right)^2} = - \dfrac{bx+2c}{\left(4ac-b^2\right)\left(ax^2+bx+c \right)} - \dfrac{b}{4ac-b^2} \int \dfrac{dx}{ax^2+bx+c}</math>
-|-
-| <math> \int \dfrac{x^2 dx}{\left(ax^2+bx+c \right)^2} = \dfrac{\left(b^2-2ac\right)x+bc}{a\left(4ac-b^2\right)\left(ax^2+bx+c\right)} + \dfrac{2c}{4ac-b^2} \int \dfrac{dx}{ax^2+bx+c}</math>
-|-
-| <math> \int \dfrac{x^{m}dx}{\left(ax^2+bx+c\right)^n} = - \dfrac{x^{m-1}}{\left(2n-m-1\right)a\left(ax^2+bx+c\right)^{n-1}} + \dfrac{\left(m-1\right)c}{\left(2n-m-1\right)a} \int \dfrac{x^{m-2}dx}{\left(ax^2+bx+c\right)^{n}} </math>
-|-
-| <math> - \dfrac{\left(n-m\right)b}{\left(2n-m-1\right)a} \int \dfrac{x^{m-1}dx}{\left(ax^2+bx+c\right)^n} </math>
-|-
-| <math> \int \dfrac{x^{2n-1}dx}{\left(ax^2+bx+c\right)^{n}} = \dfrac{1}{a} \int \dfrac{x^{2n-3}dx}{\left(ax^2+bx+c\right)^{n-1}} - \dfrac{c}{a} \int \dfrac{x^{2n-3}dx}{\left(ax^2+bx+c\right)^{n}} - \dfrac{b}{a} \int \dfrac{x^{2n-2}dx}{\left(ax^2+bx+c\right)^n} </math>
-|-
-| <math> \int \dfrac{dx}{x\left(ax^2+bx+c\right)^2} = \dfrac{1}{2c\left(ax^2+bx+c\right)} - \dfrac{b}{2c} \int \dfrac{dx}{\left(ax^2+bx+c\right)^2} + \dfrac{1}{c} \int \dfrac{dx}{x\left(ac^2+bx+c\right)} </math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(ax^2+bx+c\right)^2} = -\dfrac{1}{cx\left(ax^2+bx+c\right)} - \dfrac{3a}{c} \int \dfrac{dx}{\left(ax^2+bx+c\right)^2} - \dfrac{2b}{c} \int \dfrac{dx}{x\left(ac^2+bx+c\right)^2} </math>
-|-
-| <math> \int \dfrac{dx}{x^{m}\left(ax^2+bx+c\right)^{n}} = -\dfrac{1}{\left(m-1\right)cx^{m-1}\left(ax^2+bx+c\right)^{n-1}} - \dfrac{\left(m+2n-3\right)a}{\left(m-1\right)c} \int \dfrac{dx}{x^{m-2}\left(ax^2+bx+c\right)^{n}} </math>
-|-
-| <math> + \dfrac{\left(m+n-2\right)b}{\left(m-1\right)c} \int \dfrac{dx}{x^{m-1}\left(ac^2+bx+c\right)^{n}} </math>
-|-
-|
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math> \sqrt{ax^2 + bx + c} </math>
-|-
-| <math> Dans\quad les\quad r\acute{e}sultats\quad suivants,\quad si\quad b^2 = 4ac, \sqrt{ax^2 + bx + c} = \sqrt{a}\left(x+ b/2a\right) et\quad on\quad peut\quad utiliser\quad </math>
-|-
-| <math> les\quad r\acute{e}sultats\quad des\quad pages \quad60-61.\quad Si \quad b = 0 \quad utiliser \quad les \quad r\acute{e}sultats \quad de \quad la \quad page \quad 64. </math>
-|-
-| <math> Si \quad a \quad ou \quad c = 0,\quad utiliser \quad les\quad r\acute{e}sultats \quad des \quad pages \quad 60-61. </math>
-|-
-| <math> \int \dfrac{dx}{\sqrt{ax^2+bx+c}} =
-\begin{cases}
-\dfrac{1}{\sqrt{a}} \ln\left(2\sqrt{a}\sqrt{ax^2+bx+c}+ax+b\right)\\
--\dfrac{1}{\sqrt{-a}} \arcsin\left(\dfrac{2ax+b}{\sqrt{b^2-4ac}}\right)\quad ou\quad \dfrac{1}{\sqrt{a}} argsh\left(\dfrac{2ax+b}{\sqrt{4ac-b^2}}\right)
-\end{cases}</math>
-|-
-| <math> \int \dfrac{xdx}{\sqrt{ax^2+bx+c}} = \dfrac{\sqrt{ax^2+bx+c}}{a} - \dfrac{b}{2a} \int \dfrac{dx}{\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{x^2dx}{\sqrt{ax^2+bx+c}} = \dfrac{2ax-3b}{4a^2}\sqrt{ax^2+bx+c}+\dfrac{3b^2-4ac}{8a^2}\int\dfrac{dx}{\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{dx}{x\sqrt{ax^2+bx+c}} =
-\begin{cases}
--\dfrac{1}{\sqrt{c}}\ln\left(\dfrac{2\sqrt{c}\sqrt{ax^2+bx+c}+bx+2c}{x}\right)\\
--\dfrac{1}{\sqrt{-c}} \arcsin\left(\dfrac{bx+2c}{\left|x\right\vert\sqrt{b^2-4ac}}\right)\quad ou\quad -\dfrac{1}{\sqrt{c}}argsh\left(\dfrac{bx+2c}{\left|x\right\vert\sqrt{4ac-b^2}}\right)\end{cases} </math>
-|-
-| <math> \int \dfrac{dx}{x^2\sqrt{ax^2+bx+c}} = -\dfrac{\sqrt{ax^2+bx+c}}{cx} - \dfrac{b}{2c}\int\dfrac{dx}{x\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \sqrt{ax^2+bx+c} dx = \dfrac{\left(2ax+b\right)\sqrt{ax^2+bx+c}}{4a} + \dfrac{4ac-b^2}{8a}\int\dfrac{dx}{\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int x\sqrt{ax^2+bx+c} dx = \dfrac{\left(ax^2+bx+c\right)^{3/2}}{3a} - \dfrac{b\left(2ax+b\right)}{8a^2}\sqrt{ax^2+bx+c} - \dfrac{b\left(4ac-b^2\right)}{16a^2} \int \dfrac{dx}{\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int x^2\sqrt{ax^2+bx+c} dx = \dfrac{6ax-5b}{24a^2} \left(ax^2+bx+c\right)^{3/2} + \dfrac{5b^2-4ac}{16a^2}\int\sqrt{ax^2+bx+c} dx</math>
-|-
-| <math> \int \dfrac{\sqrt{ax^2+bx+c}}{x} dx = \sqrt{ax^2+bx+c} + \dfrac{b}{2} \int\dfrac{dx}{\sqrt{ax^2+bx+c}} + c\int\dfrac{dx}{x\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{\sqrt{ax^2+bx+c}}{x^2} dx = -\dfrac{\sqrt{ax^2+bx+c}}{x} + a\int\dfrac{dx}{\sqrt{ax^2+bx+c}} + \dfrac{b}{2}\int\dfrac{dx}{x\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{dx}{\left(ax^2+bx+c\right)^{3/2}} = \dfrac{2\left(2ax+b\right)}{\left(4ac-b^2\right)\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{xdx}{\left(ax^2+bx+c\right)^{3/2}} = \dfrac{2\left(2bx+2c\right)}{\left(b^2-4ac\right)\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int \dfrac{x^2dx}{\left(ax^2+bx+c\right)^{3/2}} = \dfrac{\left(2b^2-4ac\right)x+2bc}{a\left(4ac-b^2\right)\sqrt{ax^2+bx+c}}+\dfrac{1}{a}\int\dfrac{dx}{\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int\dfrac{dx}{x\left(ax^2+bx+c\right)^{3/2}} = \dfrac{1}{c\sqrt{ax^2+bx+c}} + \dfrac{1}{c}\int\dfrac{dx}{x\sqrt{ax^2+x+c}} - \dfrac{b}{2c}\int\dfrac{dx}{\left(ax^2+bx+c\right)^{3/2}}</math>
-|-
-| <math> \int\dfrac{dx}{x^2\left(ax^2+bx+c\right)^{3/2}} = -\dfrac{ax^2+2bx+c}{c^2x\sqrt{ax^2+bx+c}} + \dfrac{v^2-2ac}{2c^2}\int\dfrac{dx}{\left(ax^2+x+c\right)^{3/2}} - \dfrac{3b}{2c^2}\int\dfrac{dx}{x\sqrt{ax^2+bx+c}}</math>
-|-
-| <math> \int\left(ax^2+bx+c\right)^{n+1/2} dx = \dfrac{\left(2ax+b\right)\left(ax^2+bx+c\right)^{n+1/2}}{4a\left(n+1\right)} + \dfrac{\left(2n+1\right)\left(4ac-b^2\right)}{8a\left(n+1\right)}\int\left(ax^2+bx+c\right)^{n-1/2}dx</math>
-|-
-| <math> \int x\left(ax^2+bx+c\right)^{n+1/2}dx = \dfrac{\left(ax^2+bx+c\right)^{n+3/2}}{a\left(2n+3\right)} - \dfrac{b}{2a}\int \left((ax^2+bx+c\right)^{n+1/2} dx </math>
-|-
-| <math> \int \dfrac{dx}{\left(ax^2+bx+c\right)^{n+1/2}} = \dfrac{2\left(2ax+b\right)}{\left(2n-1\right)\left(4ac-b^2\right)\left(ax^2+bx+c\right)^{n-1/2}} + \dfrac{8a\left(n-1\right)}{\left(2n-1\right)\left(4ac-b^2\right)} \int \dfrac{dx}{\left(ax^2+bx+c\right)^{n-1/2}}</math>
-|-
-| <math> \int \dfrac{dx}{x\left(ax^2+bx+c\right)^{n+1/2}} = \dfrac{1}{\left(2n-1\right)c\left(ax^2+bx+c\right)^{n-1/2}} + \dfrac{1}{c} \int \dfrac{dx}{x\left(ax^2+bx+c\right)^{n-1/2}} - \dfrac{b}{2c} \int \dfrac{dx}{\left(ax^2+bx+c\right)^{n+1/2}}</math>
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math>  x^3+a^3</math>
-|-
-| <math> \int \dfrac{dx}{x^3+a^3} = \dfrac{1}{6a^2} \ln \dfrac{\left(x+a\right)^2}{x^2-ax+a^2} + \dfrac{1}{a^2\sqrt{3}} \arctan \dfrac{2x-a}{a\sqrt{3}} </math>
-|-
-| <math> \dfrac{xdx}{x^3+a^3} = \dfrac{1}{6a}\ln \dfrac{x^2-ax+a^2}{\left(x+a\right)^2} + \dfrac{1}{a\sqrt{3}} \arctan \dfrac{2x-a}{a\sqrt{3}}</math>
-|-
-| <math> \int \dfrac{x^2dx}{x^3+a^3} = \dfrac{1}{3} \ln\left(x^3+a^3\right)</math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^3+a^3\right)} = \dfrac{1}{3a^3} \ln \left(\dfrac{x^3}{x^3+a^3}\right)</math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(x^3+a^3\right)} = -\dfrac{1}{a^3x} - \dfrac{1}{6a^4} \ln \dfrac{x^2-ax+a^2}{\left(x+a\right)} - \dfrac{1}{a^4\sqrt{3}} \arctan \dfrac{2x-a}{a\sqrt{3}}</math>
-|-
-| <math> \int \dfrac{dx}{\left(x^3+a^3\right)^2} = \dfrac{x}{3a^3\left(x^3+a^3\right)} + \dfrac{1}{9a^5} \ln\dfrac{\left(x+a\right)^2}{x^2-ax+a^2} + \dfrac{2}{3a^5\sqrt{3}}\arctan\dfrac{2x-a}{a\sqrt{3}} </math>
-|-
-| <math> \int \dfrac{xdx}{\left(x^3+a^3\right)^2} = \dfrac{x^2}{3a^3\left(x^3+a^3\right)} + \dfrac{1}{18a^4} \ln \dfrac{x^2-ax+a^2}{\left(x+a\right)^2} + \dfrac{1}{3a^4\sqrt{3}} \arctan \dfrac{2x-a}{a\sqrt{3}} </math>
-|-
-| <math> \int \dfrac{x^2dx}{\left(x^3+a^3\right)} = -\dfrac{1}{3\left(x^3+a^3\right)} </math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^3+a^3\right)} = \dfrac{1}{3a^3\left(x^3+a^3\right)} + \dfrac{1}{3a^6}\ln\left(\dfrac{x^3}{x^3+a^3}\right) </math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(x^3+a^3\right)} = -\dfrac{1}{a^6x} - \dfrac{x^2}{3a^6\left(x^3+a^3\right)} - \dfrac{4}{3a^6}\int \dfrac{xdx}{x^3+a^3} </math>
-|-
-| <math> \int \dfrac{x^{m}dx}{x^3+a^3} = \dfrac{x^{m-2}}{m-2} - a^3 \int \dfrac{x^{m-3}dx}{x^3+a^3} </math>
-|-
-| <math> \int \dfrac{dx}{x^{n}\left(x^3+a^3\right)} = \dfrac{-1}{a^3\left(n-1\right)x^{n-1}} - \dfrac{1}{a^3} \int \dfrac{dx}{x^{n-3}\left(x^3+a^3\right)} </math>
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math> x^4 \pm a^4</math>
-|-
-| <math> \int \dfrac{dx}{x^4+a^4} = \dfrac{1}{4a^3\sqrt{2}}\ln\left(\dfrac{x^2+ax\sqrt{2}+a^2}{x^2-ax\sqrt{2}+a^2}\right) - \dfrac{1}{2a^3\sqrt{2}}\arctan \dfrac{ax\sqrt{2}}{x^2-a^2} </math>
-|-
-| <math> \int \dfrac{xdx}{x^4+a^4} = \dfrac{1}{2a^2} \arctan \dfrac{x^2}{a^2} </math>
-|-
-| <math> \int \dfrac{x^2da}{x^4+a^4} = \dfrac{1}{4a\sqrt{2}} \ln\left(\dfrac{x^2-ax\sqrt{2}+a^2}{x^2+ax\sqrt{2}+a^2}\right) = \dfrac{1}{2a\sqrt{2}} \arctan \dfrac{ax\sqrt{2}}{x^2-a^2} </math>
-|-
-| <math> \int \dfrac{x^3dx}{x^4+a^4} = \dfrac{1}{4} \ln \left(x^4+a^4\right) </math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^4+a^4\right)} = \dfrac{1}{4a^4}\ln\left(\dfrac{x^4}{x^4+a^4}\right) </math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(x^4+a^4\right)} = -\dfrac{1}{a^4x} - \dfrac{1}{4a^5\sqrt{2}}\ln\left(\dfrac{x^2-ax\sqrt{2}+a^2}{x^2+ax\sqrt{2}+a^2}\right) + \dfrac{1}{2a^5\sqrt{2}}\arctan\dfrac{ax\sqrt{2}}{x^2-a^2} </math>
-|-
-| <math> \int \dfrac{dx}{x^3\left(x^4+a^4\right)} = -\dfrac{1}{2a^4x^2} - \dfrac{1}{a^6} \arctan\dfrac{x^2}{a^2} </math>
-|-
-| <math> \int \dfrac{dx}{x^4-a^4} = \dfrac{1}{4a^3}\ln\left(\dfrac{x-a}{x+a}\right) - \dfrac{1}{2a^3}\arctan\dfrac{x}{a} </math>
-|-
-| <math> \int \dfrac{xdx}{x^4-a^4} = \dfrac{1}{4a^2}\ln\left(\dfrac{x^2-a^2}{x^2+a^2}\right) </math>
-|-
-| <math> \int \dfrac{x^2dx}{x^4-a^4} = \dfrac{1}{4a}\ln\left(\dfrac{x-a}{x+a}\right) + \dfrac{1}{2a}\arctan\dfrac{x}{a} </math>
-|-
-| <math> \int \dfrac{x^3dx}{x^4-a^4} = \dfrac{1}{4} \ln\left(x^4-a^4\right) </math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^3-a^4\right)} = \dfrac{1}{4a^4}\ln\left(\dfrac{x^4-a^4}{x^4}\right) </math>
-|-
-| <math> \int \dfrac{dx}{x^2\left(x^4-a^4\right)} = \dfrac{1}{a^4x} + \dfrac{1}{4a^5}\ln\left(\dfrac{x-a}{x+a}\right) + \dfrac{1}{2a^5}\arctan\dfrac{x}{a} </math>
-|-
-| <math> \int \dfrac{dx}{x^3\left(x^4-a^4\right)} = \dfrac{1}{2a^4x^2} + \dfrac{1}{4a^6}\ln\left(\dfrac{x^2-a^2}{x^2+a^2}\right) </math>
-|-
-|-
-! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Particular Integral, componant <math> x^{n} + a^{n}</math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^n+a^n\right)} = \dfrac{1}{na^{n}}\ln\dfrac{x^{n}}{x^{n}+a^{n}} </math>
-|-
-| <math> \int \dfrac{x^{n-1}}{x^{n}+a^{n}} = \dfrac{1}{n}\ln\left(x^{n}+a^{n}\right) </math>
-|-
-| <math> \int \dfrac{x^{m}}{\left(x^{n}+a^{n}\right)^{r}} = \int \dfrac{x^{m-n}dx}{\left(x^{n}+a^{n}\right)^{r-1}} - a^{n}\int\dfrac{x^{m-n}dx}{\left(x^{n}+a^{n}\right)^{r}} </math>
-|-
-| <math> \int \dfrac{dx}{x^{m}\left(x^{n}+a^{n}\right)^{r}} = \dfrac{1}{a^{n}}\int\dfrac{dx}{x^{m}\left(x^{n}+a{n}\right)^{r-1}} - \dfrac{1}{a^{n}}\int\dfrac{dx}{x^{m-n}\left(x^{n}+a^{n}\right)^{r}} </math>
-|-
-| <math> \int \dfrac{dx}{x\sqrt{x^{n}+a^{n}}} = \dfrac{1}{n\sqrt{a^{n}}}\ln\left(\dfrac{\sqrt{x^{n}+a^{n}}-\sqrt{a^{n}}}{\sqrt{x^{n}+a^{n}}+\sqrt{a^{n}}}\right) </math>
-|-
-| <math> \int \dfrac{dx}{x\left(x^{n}-a^{n}\right)} = \dfrac{1}{na^{n}}\ln\left(\dfrac{x^{n}-a^{n}}{x^{n}}\right) </math>
-|-
-| <math> \int \dfrac{x^{n-1}dx}{x^{n}-a^{n}} = \dfrac{1}{n}\ln\left(x^{n}-a^{n}\right) </math>
-|-
-| <math> \int \dfrac{x^{m}dx}{\left(x^{n}-a^{n}\right)^{r}} = a^{n}\int\dfrac{x^{m-n}dx}{\left(x^{n}-a^{n}\right)^{r}} + \int\dfrac{x^{m-n}dx}{\left(x^{n}-a^{n}\right)^{r-1}} </math>
-|-
-| <math> \int\dfrac{dx}{x^{m}\left(x^{n}-a^{n}\right)^{r}} = \dfrac{1}{a^{n}}\int\dfrac{dx}{x^{m-n}\left(x^{n}-a^{n}\right)^{r}}-\dfrac{1}{a^{n}}\int\dfrac{dx}{x^{m}\left(x^{n}-a^{n}\right)^{r-1}} </math>
-|-
-| <math> \int\dfrac{dx}{x\sqrt{x^{n}-a^{n}}} = \dfrac{2}{n\sqrt{a^{n}}}\arccos\sqrt{\dfrac{a^{n}}{x^{n}}} </math>
-|-
-| <math> \int \dfrac{x^{p-1}dx}{x^{2m}+a^{2m}} = \dfrac{1}{ma^{2m-p}} \sum_{k=1}^m \sin\dfrac{\left(2k-1\right)p\pi}{2m}\arctan\left(\dfrac{x+a\cos\left[\left(2k-1\right)\pi /2m\right]}{a\sin\left[\left(2k-1\right)\pi /2m\right]}\right) - \dfrac{1}{2ma^{2m-p}} \sum_{k=1}^m \cos\dfrac{\left(2k-1\right)p\pi}{2m}\ln\left(x^2+2ax\cos\dfrac{\left(2k-1\right)\pi}{2m} + a^2\right) </math>
-|-
-| <math> \int \dfrac{x^{p-1}dx}{x^{2m}-a^{2m}} = \dfrac{1}{2ma^{2m-p}} \sum_{k=1}^{m-1} \cos\dfrac{kp\pi}{m}\ln\left(x^2 - 2ax\cos\dfrac{k\pi}{m} + a^2\right) - \dfrac{1}{ma^{2m-p}} \sum_{k=1}^{m-1} \sin \dfrac{kp\pi}{m} \arctan\left(\dfrac{x-a\cos\left(k\pi /m\right)}{a\sin\left(k\pi/m\right)}\right) + \dfrac{1}{2ma^{2m-p}}\left\{\ln\left(x-a\right)+\left(-1\right)^{p}\ln\left(x+a\right)\right\} </math>
-|-
-| <math> \int \dfrac{x^{p-1}dx}{x^{2m+1}+a^{2m+1}} = \dfrac{2\left(-1\right)^{p-1}}{\left(2m+1\right)a^{2m-p+1}}\sum_{k=1}^m\sin\dfrac{2kp\pi}{2m+1}\arctan\left(\dfrac{x+a\cos\left[2k\pi/ \left(2m+1\right)\right]}{a\sin\left[2k\pi/ \left(2m+1\right)\right]}\right)-\dfrac{\left(-1\right)^{p-1}}{\left(2m+1\right)a^{2m-p+1}}\sum_{k=1}^m \cos\dfrac{2kp\pi}{2m+1}\ln\left(x^2+2ax\cos\dfrac{2k\pi}{2m+1}+a^2\right) + \dfrac{\left(-1\right)^{p-1}}{\left(2m+1\right)a^{2m-p+1}} </math>
-|-
-| <math> \int \dfrac{x^{p-1}dx}{x^{2m+1}-a^{2m+1}} = \dfrac{-2}{\left(2m+1\right)a^{2m-p+1}} \sum_{k=1}^m \sin \dfrac{2kp\pi}{2m+1}\arctan\left(\dfrac{x-a\cos\left[2k\pi/\left(2m+1\right)\right]}{a\sin\left[2k\pi/\left(2m+1\right)\right]}\right) + \dfrac{1}{\left(2m+1\right)a^{2m-p+1}}\sum_{k=1}^m \cos\dfrac{2kp\pi}{2m+1}\ln\left(x^2-2ax\cos\dfrac{2k\pi}{2m+1}+a^2\right) + \dfrac{\ln\left(x-a\right)}{\left(2m+1\right)a^{2m-p+1}} </math>
-|-
-|}
 ----
 [[Collective_Table_of_Formulas|Back to Collective Table of Formulas]]

Difference between revisions of "Table of indefinite integrals" - Rhea

Latest revision as of 17:06, 26 February 2015

More

Alumni Liaison