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 130: / Line 68: @@
 |<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}{\sqrt{a x +b}} =  \frac {2\sqrt{ax+b}}{a}</math>
 |-
-|<math> \int \frac {d x}{ x(ax + b)} = \frac {1}{b} \ln \left ( \frac {x}{ax +b}  \right)</math>
+|<math> \int \frac {x d x}{\sqrt{a x + b}} =  \frac {2(ax-2b)}{3a^2}\sqrt{ax+b}</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 {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^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, voir 14.59 </math>
 |-
+|<font size= 4> Click [[Table_of_indefinite_integrals_axplusb|here]] for [[Table_of_indefinite_integrals_axplusb|more integrals with ax+b]]. </font size>
+|}
+----
+==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>]]
-! 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 Including, sin ax
+----
-|-
+[[Collective_Table_of_Formulas|Back to Collective Table of Formulas]]
-|<math> \int \sin a x d x = - \frac {\cos a x }{a} </math>
-|-
-|<math> \int x \sin a x d x = \frac {\sin a x}{a^2}- \frac{x \cos a x}{a} </math>
-|-

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

Latest revision as of 17:06, 26 February 2015

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