Revision as of 13:43, 26 February 2015 by Rhea (Talk | contribs)


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 $

Back to Table of Indefinite Integrals

Back to Collective Table of Formulas

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang