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Indefinite Integrals
1 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 $
$ \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 \ \text{and} \ a \neq 1 $
$ \int \sin u\ d u = - \cos u $
$ \int \cos u\ d u = \sin u $
$ \int \tan u\ d u = - \ln {\cos u} $
$ \int \cot u\ d u = \ln {\sin u} $
$ \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 )}} $
$ \int \frac{d u}{\sin u} = \ln { \left ( \frac{1}{\sin u} - \cot u \right )} = \ln{\tan { \frac{u}{2}}} $
$ \int \frac{d u}{\cos ^2 u} = \tan u $
$ \int \frac{d u}{\sin ^2 u} = - \cot u $
$ \int \tan ^2 u \ d u = \tan u - u $
$ \int \cot ^2 u \ d u = - \cot u - u $
$ \int \sin ^2 u \ d u= \frac{u}{2} - \frac{\sin {2 u}}{4} = \frac{1}{2}\left( u - \sin u \cos u \right ) $
$ \int \frac {1}{\cos u} \tan u \ d u = \frac{1}{\cos u} $
$ \int \frac {1}{\sin u} \cot u \ d u = - \frac{1}{\sin u} $
$ \int \sinh u \ d u = \coth u $
$ \int \cosh u \ d u = \sinh u $
$ \int \tanh u \ d u = \ln \cosh u $
$ \int \coth u \ d u = \ln \sinh u $
$ \int \frac {1}{\operatorname{ch}\ u} \ d u = \arcsin{\left ( \operatorname{th}\,u \right )} \qquad or \ 2 arc \ th \ e^u $
$ \int \frac {1}{\operatorname{sh}\ u} \ d u = \ln \operatorname{th}\,\frac{2}{2} \qquad or \ - \operatorname{Arg coth} \ e^u $
$ \int \frac {1}{\operatorname{ch^2}\ u} \ d u = \operatorname{th}\,u $
$ \int \frac {1}{\operatorname{sh^2}\ u} \ d u = - \operatorname{coth}\,u $
$ \int \operatorname{th^2}\ u \ d u = u - \operatorname{th}\,u $
$ \int \operatorname{coth^2}\ u \ d u = u - \operatorname{coth}\,u $
$ \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 ) $
$ \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 ) $
$ \int \frac{\operatorname th \ u}{\operatorname ch \ u} \ d u = - \frac {1}{\operatorname ch \, u } $
$ \int \frac{\operatorname coth \ u}{\operatorname sh \ u} \ d u = - \frac {1}{\operatorname sh \, u } $
$ \int \frac{d u}{u^2 + a^2} = \frac {1}{a}\arctan \frac{u}{a} $
$ \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 $
$ \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 $
$ \int \frac{d u}{\sqrt{a^2 - u^2}} = \arcsin \frac{u}{a} $
$ \int \frac{d u}{\sqrt{u^2 + a^2}} = \ln { \left ( u + \sqrt {u^2+a^2} \right ) } \qquad or \ \operatorname{argth} \ \frac{u}{a} $
$ \int \frac{d u}{\sqrt{u^2 - a^2}} = \ln { \left ( u + \sqrt {u^2-a^2} \right ) } $
$ \int \frac{d u}{u \sqrt{u^2 - a^2}} = \frac {1}{a} \arccos \left | \frac{a}{u} \right | $
$ \int \frac{d u}{u \sqrt{u^2 + a^2}} = - \frac {1}{a} \ln \left ( \frac{a + \sqrt{u^2 + a^2}}{u} \right ) $
$ \int \frac{d u}{u \sqrt{a^2 - u^2}} = - \frac {1}{a} \ln \left ( \frac{a + \sqrt{a^2 - u^2}}{u} \right ) $
$ \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 $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett