Integrals: Difference between revisions

Revision as of 15:23, 24 May 2026

$\int u d v = u v - \int v d u$
$\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C, (n \neq - 1)$
$\int \frac{d u}{u} = \ln | u | + C$
$\int e^{u} d u = e^{u} + C$
$\int a^{u} d u = \frac{a^{u}}{\ln a} + C$
$\int \sin u d u = - \cos u + C$
$\int \cos u d u = \sin u + C$
$\int \sec^{2} u d u = \tan u + C$
$\int \csc^{2} u d u = - \cot u + C$
$\int \sec u \tan u d u = \sec u + C$
$\int \csc u \cot u d u = - \csc u + C$
$\int \tan u d u = \ln | \sec u | + C$
$\int \cot u d u = \ln | \sin u | + C$
$\int \sec u d u = \ln | \sec u + \tan u | + C$
$\int \csc u d u = \ln | \csc u - \cot u | + C$
$\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \sin^{- 1} \frac{u}{a} + C, (a > 0)$
$\int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \tan^{- 1} \frac{u}{a} + C$
$\int \frac{d u}{u \sqrt{u^{2} - a^{2}}} = \frac{1}{a} \sec^{- 1} \frac{u}{a} + C$
$\int \frac{d u}{a^{2} - u^{2}} = \frac{1}{2 a} \ln | \frac{u + a}{u - a} | + C$
$\int \frac{d u}{u^{2} - a^{2}} = \frac{1}{2 a} \ln | \frac{u - a}{u + a} | + C$

@@ Line 2: / Line 2: @@
 == Integral Forms & Antiderivatives ==
 === Basic Forms ===
 # <math>\int u \, dv = uv - \int v \, du</math><br>
 # <math>\int u^n \, du = \frac{u^{n+1}}{n+1} + C, \quad (n \neq -1)</math><br>
@@ Line 23: / Line 22: @@
 # <math>\int \frac{du}{a^2 - u^2} = \frac{1}{2a} \ln \left| \frac{u+a}{u-a} \right| + C</math><br>
 # <math>\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C</math><br>
+<br>
+=== Forms Involving <math>\sqrt{a}</math> ===
+== To Do ==
+*Antiderivative DB search system.