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 1: / Line 1: @@
+[[Category:Formulas]]
+[[Category:integral]]
+<center><font size= 4>
+'''[[Collective_Table_of_Formulas|Collective Table of Formulas]]'''
+</font size>
+'''Indefinite Integrals'''
+click [[Collective_Table_of_Formulas|here]] for [[Collective_Table_of_Formulas|more formulas]]
+</center>
+----
 {|
 |-
-! 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 Infinite 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" | General Rules
 |-
 |<math> \int a d x =  a x </math>
@@ Line 15: / Line 27: @@
 |<math> \int f ( a x ) d x = \frac{1}{a} \int f ( u ) d u </math>
 |-
-|<math> \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 ) </math>
+|<math> \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 ) </math>
 |-
 |<math> \int u^n d u = \frac{u^{n+1}}{n+1} \qquad n \neq -1 </math>
 |-
-|<math> \int \frac{d u}{u} =  \ln u \ ( if \ u > 0 ) \ or \ln {-u} \ ( if \ u < 0 ) = \ln \left | u \right | </math>
+|<math> \int \frac{d u}{u} =  \ln u \ ( if \ u > 0 ) \ \text{or} \ln {-u} \ ( \text{if} \ u < 0 ) = \ln \left | u \right | </math>
 |-
 |<math> \int e^u d u = e^u </math>
 |-
-|<math> \int a^u d u = \int e^{u \ln a} d u = \frac{e^{u \ln a}}{\ln a} = \frac{a^u}{\ln a} \qquad a > 0 \ and \ a \neq 1</math>
+|<font size= 4> Click [[Table_of_indefinite_integrals_general_rules|here]] for [[Table_of_indefinite_integrals_general_rules|more general rules]]. </font size>
 |-
-|<math> \int \sin u \ d u = - \cos u </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" | Transformations of the independent variable
-|-
-|<math> \int \cos u \ d u =  \sin u </math>
-|-
-|<math> \int \tan u \ d u =  - \ln {\cos u} </math>
-|-
-|<math> \int \cot u \ d u =   \ln {\sin u} </math>
-|-
-|<math> \int \frac{d u}{\cos u} =   \ln { \left ( \frac{1}{\cos u} + \tan u \right )}  = \ln{\tan   {\left ( \frac{u}{2}+\frac{\pi}{4}\right )}} </math>
-|-
-|<math> \int \frac{d u}{\sin u} =   \ln { \left ( \frac{1}{\sin u} - \cot u \right )}  = \ln{\tan   { \frac{u}{2}}} </math>
-|-
-|<math> \int \frac{d u}{\cos ^2 u} = \tan u </math>
-|-
-|<math> \int \frac{d u}{\sin ^2 u} = - \cot u </math>
-|-
-|<math> \int \tan ^2 u \ d u =  \tan u - u</math>
-|-
-|<math> \int \cot ^2 u \ d u =  - \cot u - u</math>
-|-
-|<math> \int \sin ^2 u  \ d u=  \frac{u}{2} - \frac{\sin {2 u}}{4} = \frac{1}{2}\left( u - \sin u \cos u \right )</math>
-|-
-|<math> \int \frac {1}{\cos  u} \tan u \ d u =  \frac{1}{\cos u}</math>
-|-
-|<math> \int \frac {1}{\sin  u} \cot u \ d u = - \frac{1}{\sin u}</math>
-|-
-|<math> \int \operatorname{sh}\,u  \ d u =  \operatorname{ch}\,u</math>
-|-
-|<math> \int \operatorname{ch}\,u  \ d u =  \operatorname{sh}\,u</math>
-|-
-|<math> \int \operatorname{th}\,u  \ d u =  \ln \operatorname{ch}\,u</math>
-|-
-|<math> \int \operatorname{coth}\,u  \ d u = \ln \operatorname{sh}\,u</math>
-|-
-|<math> \int \frac {1}{\operatorname{ch}\ u}  \ d u = \arcsin{\left ( \operatorname{th}\,u \right )} \qquad 2 arc th  e^u</math>
-|-
-|<math> \int \frac {1}{\operatorname{sh}\ u}  \ d u = \ln \operatorname{th}\,\frac{2}{2} \qquad - Arg \coth e^u</math>
-|-
-|<math> \int \frac {1}{\operatorname{ch^2}\ u}  \ d u =  \operatorname{th}\,u  </math>
-|-
-|<math> \int \frac {1}{\operatorname{sh^2}\ u}  \ d u = - \operatorname{coth}\,u  </math>
-|-
-|<math> \int \operatorname{th^2}\ u \ d u = u - \operatorname{th}\,u  </math>
-|-
-|<math> \int \operatorname{coth^2}\ u \ d u = u - \operatorname{coth}\,u  </math>
-|-
-|<math> \int \operatorname{sh^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} - \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u  - u \right )</math>
-|-
-|<math> \int \operatorname{ch^2}\ u \ d u = \frac {\operatorname{sh}\,{2 u}}{4} + \frac{u}{2}=\frac{1}{2}\left ( \operatorname{sh}\,u \ \operatorname{ch}\,u  + u \right )</math>
-|-
-|<math> \int \frac{\operatorname th \ u}{\operatorname ch \ u} \ d u = - \frac {1}{\operatorname ch \, u }</math>
-|-
-|<math> \int \frac{\operatorname coth \ u}{\operatorname sh \ u} \ d u = - \frac {1}{\operatorname sh \, u }</math>
-|-
-|<math> \int \frac{d u}{u^2 + a^2} =  \frac {1}{a}\arctan \frac{u}{a}</math>
-|-
-|<math> \int \frac{d u}{u^2 - a^2} =  \frac {1}{2 a}\ln \left ( \frac{u-a}{u+a} \right ) = -\frac{1}{a} \operatorname{argcoth}\frac{u}{a} \qquad u^2 > a^2 </math>
-|-
-|<math> \int \frac{d u}{a^2 - u^2} =  \frac {1}{2 a}\ln \left ( \frac{a+u}{a-u} \right ) = \frac{1}{a} \operatorname{argth}\ \frac{u}{a} \qquad u^2 < a^2 </math>
-|-
-|<math> \int \frac{d u}{\sqrt{a^2 - u^2}} = \arcsin  \frac{u}{a}  </math>
-|-
-|<math> \int \frac{d u}{\sqrt{u^2 + a^2}} = \ln { \left ( u + \sqrt {u^2+a^2} \right ) } \qquad \operatorname{argth} \ \frac{u}{a}  </math>
-|-
-|<math> \int \frac{d u}{\sqrt{u^2 - a^2}} = \ln { \left ( u + \sqrt {u^2-a^2} \right ) } </math>
-|-
-|<math> \int \frac{d u}{u \sqrt{u^2 - a^2}} = \frac {1}{a} \arccos \left | \frac{a}{u} \right |  </math>
-|-
-|<math> \int \frac{d u}{u \sqrt{u^2 + a^2}} = - \frac {1}{a} \ln \left ( \frac{a + \sqrt{u^2 + a^2}}{u} \right )   </math>
-|-
-|<math> \int \frac{d u}{u \sqrt{a^2 - u^2}} = - \frac {1}{a} \ln \left ( \frac{a + \sqrt{a^2 - u^2}}{u} \right )   </math>
-|-
-|<math> \int f^{(n)} \ g d x  =f^{(n-1)} \ g - f^{(n-2)} \ g' + f^{(n-3)} \ g'' - \cdot \cdot \cdot \ (-1)^n \int fg^{(n)} d x  </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" | Important Transformations
 |-
 |<math> \int F( a x + b) d x =\frac{1}{a} \int F( u) d u  \qquad  u = a x + b</math>
@@ Line 122: / 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" | Particular Integral, component 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 129: / Line 67: @@
 |-
 |<math> \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)</math>
-|-
-|<math> \int \frac {x^3 d x}{ ax + b} = \frac {(ax+b)^3}{3a^4} - \frac {3b(ax+b)^2 }{2a^4} + \frac{3b^2(ax+b)}{a^4} - frac{b^3}{a^3}\ln (ax +b)</math>
-|-
-|<math> \int \frac {d x}{ x(ax + b)} = \frac {1}{b} \ln \left ( \frac {x}{ax +b}  \right)</math>
-|-
-|<math> \int \frac {d x}{ x^2(ax + b)} = - \frac {1}{b x} + \frac {a}{b^2} \ln \left ( \frac {ax +b}{x}  \right)</math>
-|-
-|<math> \int \frac {d x}{ x^3(ax + b)} =  \frac {2 a x - b}{2 b^2 x^2} + \frac {a^2}{b^3} \ln \left ( \frac {x}{ax+b}  \right)</math>
-|-
-|<math> \int \frac {d x}{(ax + b)^2} =  \frac {-1}{a(ax+b)} </math>
-|-
-|<math> \int \frac {x d x}{(ax + b)^2} =  \frac {b}{a^2(ax+b)} + \frac {1}{a^2} \ln (ax+b) </math>
-|-
-|<math> \int \frac {x^2 d x}{(ax + b)^2} =  \frac {ax+b}{a^3} - \frac{b^2}{a^3(ax+b)} - \frac {2b}{a^3} \ln (ax+b) </math>
-|-
-|<math> \int \frac {x^3 d x}{(ax + b)^2} =  \frac {(ax+b)^2}{2a^4} - \frac {3b(ax+b)}{a^4} +\frac{b^3}{a^4(ax+b)} + \frac {3b^2}{a^4} \ln (ax+b) </math>
-|-
-|<math> \int \frac {d x}{x(ax + b)^2} =  \frac {1}{b(ax+b)} + \frac {1}{b^2} \ln \left ( \frac{x}{ax+b} \right )  </math>
-|-
-|<math> \int \frac {d x}{x^2(ax + b)^2} =  \frac {-a}{b^2(ax+b)} - \frac {1}{b^2x} + \frac {2a}{b^3} \ln \left ( \frac {ax+b}{x} \right )  </math>
-|-
-|<math> \int \frac {d x}{x^3(ax + b)^2} = - \frac {(ax+b)^2}{2b^4x^2} + \frac {3 a(ax+b)}{b^4x} - \frac {a^3 x}{b^4(ax+b)} - \frac{3a^2}{b^4} \ln \left ( \frac {ax+b}{x} \right )  </math>
-|-
-|<math> \int \frac {d x}{(ax + b)^3} =  \frac {-1}{2(ax+b)^2}  </math>
-|-
-|<math> \int \frac {x d x}{(ax + b)^3} =  \frac {-1}{a^2(ax+b)} + \frac {b}{2a^2(ax+b)^2}  </math>
-|-
-|<math> \int \frac {x^2 d x}{(ax + b)^3} =  \frac {2b}{a^3(ax+b)} - \frac {b^2}{2a^3(ax+b)^2} + \frac {1}{a^3} \ln (ax+b)  </math>
-|-
-|<math> \int \frac {x^3 d x}{(ax + b)^3} =  \frac {x}{a^3} - \frac {3b^2}{a^4(ax+b)} + \frac {b^3}{2a^4(ax+b)^2} - \frac {3b}{a^4} \ln (ax+b)  </math>
-|-
-|<math> \int \frac {d x}{x(ax + b)^3} =  \frac {a^2x^2}{2b^3(ax+b)^2} - \frac {2ax}{b^3(ax+b)} - \frac {1}{b^3} \ln \left( \frac{ax+b}{x} \right)  </math>
-|-
-|<math> \int \frac {d x}{x^2(ax + b)^3} =  \frac {-a}{2b^2(ax+b)^2} - \frac {2a}{b^3(ax+b)} - \frac {1}{b^3x} + \frac {3a}{b^4} \ln \left( \frac{ax+b}{x} \right)  </math>
-|-
-|<math> \int \frac {d x}{x^3(ax + b)^3} =  \frac {a^4x^2}{2b^5(ax+b)^2} - \frac {4a^3x}{b^5(ax+b)} - \frac {(ax+b)^2}{2b^5x2} - \frac {6a^2}{b^5} \ln \left( \frac{ax+b}{x} \right)  </math>
-|-
-|<math> \int (a x +b)^n  d x =  \frac {(ax+b)^{n+1} }{(n+1)a}. \qquad n =-1 </math>
-|-
-|<math> \int x (a x +b)^n  d x =  \frac {(ax+b)^{n+2} }{(n+2)a^2} - \frac {b(ax+b)^{n+1}}{(n+1)a^2}, \qquad n \neq -1,-2 </math>
-|-
-|<math> \int x^2 (a x + b)^n  d x =  \frac {(ax+b)^{n+3} }{(n+3)a^3} - \frac {2b(ax+b)^{n+2}}{(n+2)a^3} + \frac {b^2(ax+b)^{n+1}}{(n+1)a^3} \qquad n = -1,-2, -3 </math>
-|-
-|<math> \int x^m (a x + b)^n  d x =
-\begin{cases}
- \frac {x^{m+1}(ax+b)^n}{m + n + 1} + \frac {n b}{m + n+ 1} \int x^m (ax+b)^{n-1} d x \\
- \frac {x^m(ax+b)^{n+1}}{(m + n + 1)a} - \frac {m b}{(m + n+ 1)a} \int x^{m-1} (ax+b)^{n} d x \\
- \frac {- x^{m+1}(ax+b)^{n+1}}{(n + 1)b} + \frac {m+ n+ 2 }{(n+ 1)b} \int x^m (ax+b)^{n+1} d x
-\end{cases} </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, component \sqrt{ax +b}
 |-
 |<math> \int \frac {d x}{\sqrt{a x +b}} =  \frac {2\sqrt{ax+b}}{a}</math>
@@ Line 188: / Line 74: @@
 |<math> \int \frac {x^2 d x}{\sqrt{a x + b}} =  \frac {2(3a^2x^2-4abx + 8b^2)}{15a^3}\sqrt{ax+b}</math>
 |-
-|<math> \int \frac {d x}{x \sqrt {ax+b}} =
+|<font size= 4> Click [[Table_of_indefinite_integrals_axplusb|here]] for [[Table_of_indefinite_integrals_axplusb|more integrals with ax+b]]. </font size>
-\begin{cases}
+|}
- \frac {1}{b} \ln \left ( \frac {\sqrt {ax+b} - \sqrt {b}}{\sqrt {ax+b} + \sqrt {b}} \right )  \\
+----
- \frac {2}{\sqrt {-b}} \arctan \sqrt { \frac {ax+b}{- b}}  \\
+==More==
+* [[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>]]
+* [[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>]]
+*[[Table_of_indefinite_integrals_cotangent|Integrals with cotangent (cot x)]]
+*[[Table_of_indefinite_integrals_oneovercosine|Integrals with 1/cos x]]
+*[[Table_of_indefinite_integrals_inversecircularfunctions|Integrals with arccos, arcsin, arctan, arc cot]]
+*[[Table_of_indefinite_integrals_exponential|Integrals with <math>e^x</math>]]
+*[[Table_of_indefinite_integrals_log|Integrals with <math>\ln x</math>]]
+*[[Table_of_indefinite_integrals_sh|Integrals with hyperbolic sine (sh x)]]
+*[[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 (coth x)]]
+*[[Table_of_indefinite_integrals_x_inverse|Integrals with 1/x]]
+*[[Table_of_indefinite_quadratic|Integrals with <math>ax^2+bx+c</math>]]
-\end{cases} </math>
+----
-|-
+[[Collective_Table_of_Formulas|Back to Collective Table of Formulas]]
-|<math> \int \frac { d x}{x ^2 \sqrt{a x + b}} = - \frac {\sqrt{ax+b}}{b x} - \frac {a}{2 b} \int \frac {d x}{x \sqrt {ax + b}}</math>
-|-
-|<math> \int \sqrt{a x + b} \ d x =  \frac {2 \sqrt{(ax+b)3}}{3 a}</math>
-|-
-|<math> \int x \sqrt{a x + b} \ d x =  \frac {2(3ax-2b)}{15a^2}\sqrt{(ax+b)^3}</math>
-|-
-|<math> \int x^2 \sqrt{a x + b} \ d x =  \frac {2(15a^2x^2-12abx + 8b^2)}{105a^3}\sqrt{(ax+b)^3}</math>
-|-
-|<math> \int \frac {\sqrt {ax+b}}{x} \ d x = 2 \sqrt {ax+b} + b \ \int \frac {d x}{x \sqrt {ax + b}} </math>
-|-
-|<math> \int \frac {\sqrt {ax+b}}{x^2} \ d x = - \frac {\sqrt {ax+b}}{x} +  \frac {a}{2} \int \frac {d x}{x \sqrt {ax + b}} </math>
-|-
-|<math> \int \frac {x^m}{\sqrt{ ax+b}} d x = \frac {2x^m \sqrt {ax+b}}{(2m+1)a} - \frac {2mb}{(2m+1)a} \int \frac {x^{m-1}}{\sqrt {ax+b}} d x</math>
-|-
-|<math> \int \frac {d x}{x^m \sqrt{ax+b}} =- \frac {\sqrt{ax+b}}{(m-1)bx^{m-1}} - \frac {(2m-3)a}{(2m-2)b} \int \frac {d x}{x^{m-1} \sqrt{ax+b}}</math>
-|-
-|<math> \int x^m \sqrt {ax+b} \ d x = \frac{2x^m}{(2m+3)a}(a+b)^{\frac{3}{2}} -\frac {2mb}{(2m+3)a} \int x^{m-1} \sqrt{ax+b} \ d x</math>
-|-
-|<math>\int \frac {\sqrt {ax+b}}{x^m} d x = - \frac {\sqrt {ax+b}}{(m-1)x^{m-1}} + \frac {a}{2(m-1)} \int \frac {d x}{x^{m-1} \sqrt {ax+b}} </math>
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-|<math>\int \frac {\sqrt {ax+b}}{x^m} d x =  \frac {-(ax+b)^{3/2}}{(m-1)bx^{m-1}} - \frac {(2m-5)a}{(2m-2)b} \int \frac {\sqrt {ax+b}}{x^{m-1}} d x </math>
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-|<math> \int (ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+2)/2}}{a(m+2)}</math>
-|-
-|<math> \int x(ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+4)/2}}{a^2(m+4)} - \frac {2b(ax+b)^{(m+2)/2}}{a^2(m+2)}</math>
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-|<math> \int x^2(ax+b)^{m/2} d x = \frac {2(ax+b)^{(m+6)/2}}{a^3(m+6)} - \frac {4b(ax+b)^{(m+4)/2}}{a^3(m+4)}+ \frac {2b^2(ax+b)^{(m+2)/2}}{a^3(m+2)}</math>
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-|<math> \int \frac {(ax+b)^{m/2}}{x} d x =\frac {2(ax+b)^{m/2}}{m} + b \ \int \frac {(ax+b)^{(m-2)/2}}{x} d x</math>
-|-
-|<math> \int \frac {(ax+b)^{m/2}}{x^2} d x = - \frac {(ax+b)^{(m+2)/2}}{bx} + \frac {ma}{2b} \ \int \frac {(ax+b)^{m/2}}{x} d x</math>
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-|<math> \int \frac {d x}{x(ax+b)^{m/2}} d x =  \frac {2}{(m-2)b(ax+b)^{(m-2)/2}} + \frac {1}{b} \ \int \frac {d x}{x(ax+b)^{(m-2)/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, component ax + b ET px + q
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-|<math> \int \frac {d x}{(ax+b)(px+q)} = \frac {1}{bp-aq} \ln \left ( \frac {px+q}{ax+b} \right )</math>
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-|<math> \int \frac {x d x}{(ax+b)(px+q)} = \frac {1}{bp-aq} \left \{ \frac{b}{a} \ln { (ax+b)} - \frac{q}{p} \ln{(px+q)} \right \}</math>
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-|<math> \int \frac { d x}{(ax+b)^2(px+q)} = \frac {1}{bp-aq} \left \{ \frac {1}{ax+b} + \frac {p}{bp- aq} \ln  \left ( \frac {px+q}{ax +b} \right ) \right \}</math>
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-|<math> \int \frac {x d x}{(ax+b)^2(px+q)} = \frac {1}{bp-aq} \left \{ \frac {q}{bp- aq} \ln  \left ( \frac {ax+b}{px +q} \right ) - \frac {b}{a(ax+b)} \right \}</math>
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-|<math> \int \frac {x^2 d x}{(ax+b)^2(px+q)} = \frac {b^2}{(bp -aq)a^2(ax+b)} + \frac {1}{(bp-aq)^2} \left \{ \frac {q^2}{p} \ln (px+q) + \frac {b(bp-2aq)}{a^2} \ln (ax+b) \right \}</math>
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-|<math> \int \frac { d x }{(ax+b)^m (px+q)^n } = \frac {-1}{(n-1)(bp-aq)} \left \{ \frac {1}{(ax+b)^{m-1}(pz+q)^{n-1} } + a(m + n -2) \ \int \frac {d x}{(ax+b)^m(px+q)^{n-1} } \right \} </math>
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-|<math> \int \frac {ax+b}{px+q} d x = \frac {ax}{p} + \frac {bp -aq}{p^2} \ln (px+q)</math>
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-|<math> \int \frac {(ax+b)^m}{(px+q)^n} d x =
-\begin{cases}
- \frac {-1}{(n-1)(bp-aq)} \left \{ \frac {(ax+b)^{m+1}}{(px+q)^{n-1}} + (n-m-2)a \int \frac {(ax+b)^m}{(px+q)^{n-1}} d x \right \}  \\
- \frac {-1}{(n-m-1)p} \left \{ \frac {(ax+b)^m}{(px+q)^{n-1}} + m(bp-aq) \int \frac{(ax+b)^{m-1}} {(px+q)^n} d x \right \}  \\
- \frac{-1}{(n-1)p} \left \{ \frac{(ax+b)^m}{(px+q)^{n-1}} - ma \int \frac{(ax+b)^{m-1}}{(px+q)^{n-1}} d x \right \}
-\end{cases} </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, component sqrt{ax +b} ET px+q
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-|<math> \int \frac {px+q}{\sqrt {ax+b}} d x = \frac {2(apx+3aq-2bp)}{3a^2} \sqrt {ax+b}</math>
-|-
-|<math> \int \frac {d x}{(px+q) \sqrt {ax+b}} =
-\begin{cases}
- \frac{1}{ \sqrt {bp-aq} \sqrt {p} } \ln \left ( \frac {\sqrt{p(ax+b)} -\sqrt {bp-aq} }{\sqrt{p(ax+b)} + \sqrt {bp-aq} } \right ) \\
-\frac {2}{\sqrt {aq-bp} \sqrt {p}} \arctan \sqrt{ \frac {p(ax+b)}{aq-bp} }
-\end{cases}</math>
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-|<math> \int \frac {\sqrt {ax+b}}{px+q} \ d x =
-\begin{cases}
-\frac{2 \sqrt{ax+b}}{p} \ + \ \frac {\sqrt {bp-aq}}{p \sqrt{p}} \ln \left ( \frac {\sqrt {p(ax+b)} - \sqrt {bp-aq}}{\sqrt {p(ax+b)} + \sqrt {bp-aq}} \right ) \\
-\frac {2 \sqrt {ax+b}}{p} \  - \ \frac {2 \sqrt{aq-bp}}{p \sqrt{p}} \arctan \sqrt { \frac {p(ax+b)}{aq-bp}} \\
-\end{cases}</math>
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-|<math> \int (px+q)^n \sqrt {ax+b} \ d x = \frac{2(px+q)^{n+1} \sqrt {ax+b}}{(2n+3)p} \ + \  \frac {bp-aq}{(2n+3)p} \int \frac {(px+q)^n}{\sqrt {ax+b}} d x</math>
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-|<math> \int \frac {d x}{ (px+q)^n \sqrt {ax+b}} = \frac {\sqrt {ax+b}}{(n-1)(aq-bp)(px+q)^{n-1}} + \frac {(2n-3)a}{2(n-1)(aq-bp)} \int \frac {d x }{(px+q)^{n-1} \sqrt{ax+b}} </math>
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-|<math>\int \frac {(px+q)^n}{\sqrt {ax+b}} d x = \frac {2(px+q)^n \sqrt{ax+b}}{)2n+1)a} + \frac {2n(aq-bp)}{(2n+1)a} \int \frac {(px+q)^{n-1} d x}{\sqrt{ax+b}} </math>
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-|<math> \int \frac {\sqrt{ax+b}}{(px+q)^n} d x = \frac {- \sqrt {ax+b}}{(n-1)p(px+q)^n-1} + \frac{a}{2(n-1)p} \int \frac {d x}{(px+q)^{n-1} \sqrt {ax+b}} </math>
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-! 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 Componant sqrt(ax+b) ET sqrt (px+q)
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-|<math>\int \frac {d x}{\sqrt {(ax+b)(px+q)}} =
-\begin{cases}
-\frac{2}{\sqrt{ap}} \ln { \left ( \sqrt {a(px+q)}+ \sqrt{p(ax+b)} \right )} \\
-\frac {2}{\sqrt { -ap}} \arctan \ \sqrt { \frac {-p(ax+b)} {a(px+q)} }  \\
-\end{cases} </math>
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-|<math>\int \frac {x d x}{\sqrt {(ax+b)(px+q)}} = \frac {\sqrt{(ax+b)(px+q)}}{ap} - \frac{bp+aq}{2ap} \int \frac {d x}{\sqrt{(ax+b)(px+q)}}</math>
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-|<math>\int \sqrt {(ax+b)(px+q)} dx = \frac {2apx + bp +aq}{4ap} \sqrt {(ax+b)(px+q)} - \frac {(bp-aq)^2}{8ap} \int \frac{d x}{\sqrt{(ax+b)(px+q)}}</math>
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-! 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" | 19 Integrals Component sin ax
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-|<math> \int \sin a x d x = - \frac {\cos a x }{a} </math>
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-|<math> \int x \sin a x d x = \frac {\sin a x}{a^2}- \frac{x \cos a x}{a} </math>
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-|<math> \int x^2 \sin a x d x = \frac {2 x}{a^2} \sin a x + \left ( \frac {2}{a^3} - \frac {x^2}{a} \right)\cos a x </math>
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-|<math> \int x^3 \sin a x d x = \left( \frac {3 x^2}{a^2} - \frac{6}{a^4}\right)\sin a x + \left ( \frac {6x}{a^3} - \frac {x^3}{a} \right)\cos a x </math>
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-|<math> \int \frac {\sin a x}{x} d x = a x - \frac {(a x)^3}{3 \cdot 3!} + \frac {(a x)^5}{5 \cdot 5!} - \cdot \cdot \cdot </math>
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-|<math> \int \frac {\sin a x}{x^2} d x = - \frac {\sin a x}{x} + a \int \frac {\cos a x}{x}dx \qquad \left( \text {Voir} \text{ } 14.373 \right) </math>
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-|<math> \int \frac {d x}{\sin a x} = \frac {1}{a} \ln \left( \frac {1}{\sin a x} - \cot a x \right) = \frac {1}{a} \ln {\tan \frac{a x}{2}}  </math>
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-|<math> \int \frac {x d x}{\sin a x} = \frac {1}{a^2} \left \{a x + \frac{(ax)^3}{18}+ \frac {7(ax)^5}{1800} + \cdot \cdot \cdot + \frac {2(2^{2n-1}-1)Bn(ax)^{2n+1}}{(2n+1)!} + \cdot\cdot\cdot \right \} </math>
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-|<math> \int \sin ^2 a x d x = \frac {x}{2}- \frac{\sin 2 a x}{4a} </math>
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-|<math> \int x \sin ^2 a x d x = \frac {x^2}{4}- \frac{x \sin 2 a x}{4a} - \frac {\cos 2 a x}{8a^2}</math>
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-|<math> \int \sin ^3 a x d x = -\frac {\cos a x}{a}- \frac{\cos^3 a x}{3a} </math>
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-|<math> \int \sin ^4 a x d x = \frac {3x}{8}- \frac{\sin 2 a x}{4a} + \frac {\sin4ax}{32a} </math>
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-|<math> \int \frac {d x}{\sin^2 a x} = -\frac {1}{a} \cot a x </math>
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-|<math> \int \frac {d x}{\sin^3 a x} = -\frac {\cos ax}{2a \sin^2 ax} + \frac{1}{2a}\ln \tan \frac{ax}{2} </math>
-|-
-|<math> \int \sin px \sin q x d x = \frac {\sin (p-q)x}{2(p-q)} - \frac{\sin(p+q)x}{2(p+q)} \qquad \left( \text {Si} \text{ } p=\pm q, \text {voir}\text{ }14.368 \right)</math>
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-|<math> \int \frac {d x}{1-\sin a x} = \frac {1}{a} \tan   {\left ( \frac{\pi}{4}+\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {x d x}{1-\sin a x} = \frac {x}{a} \tan   {\left ( \frac{\pi}{4}+\frac{ax}{2}\right )} +\frac{2}{a^2}\ln\sin{\left ( \frac{\pi}{4}-\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {d x}{1+\sin a x} = -\frac {1}{a} \tan   {\left ( \frac{\pi}{4}-\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {x d x}{1+\sin a x} = -\frac {x}{a} \tan   {\left ( \frac{\pi}{4}-\frac{ax}{2}\right )} +\frac{2}{a^2}\ln\sin{\left ( \frac{\pi}{4}+\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {d x}{(1-\sin a x)^2} = \frac {1}{2a} \tan   {\left ( \frac{\pi}{4}+\frac{ax}{2}\right )} +\frac{1}{6a}\tan^3{\left ( \frac{\pi}{4}+\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {d x}{(1+\sin a x)^2} = -\frac {1}{2a} \tan   {\left ( \frac{\pi}{4}-\frac{ax}{2}\right )} -\frac{1}{6a}\tan^3{\left ( \frac{\pi}{4}-\frac{ax}{2}\right )} </math>
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-|<math> \int \frac {dx}{(p+q \sin ax)}=
-\begin{cases}
-\frac{2}{a \sqrt {p^2-q^2}} \arctan \frac {p\tan \frac{1}{2}ax + q}{\sqrt{p^2-q^2}} \\
-\frac{1}{a\sqrt{q^2-p^2}} \ln \left( \frac{p \tan \frac{1}{2} ax + q-\sqrt{q^2-p^2}}{p \tan \frac{1}{2}ax+q+\sqrt{q^2-p^2}} \right)
-\end{cases}
-\text{      Si} \text{ }p= \pm q \text{voir}\text{ }14.354 \text{et} \text{ } 14.356 </math>
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-|<math> \int \frac {d x}{(p+q \sin a x)^2}=\frac{q \cos ax}{a(p^2-q^2)(p+q \sin ax)}+\frac{p}{p^2-q^2}\int\frac{dx}{p+q \sin ax} \text{      Si} \text{ }p= \pm q \text{voir}\text{ }14.358 \text{et} \text{ } 14.359 </math>

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

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