Basic Anti-Derivatives

Question	Answer
$\int k \, \textrm{d}x =$	$kx +C$
$\int \, \textrm{d}x =$	$x +C$
$\int 0 \, \textrm{d}x =$	$C$
$\int k \, f(x) \, \textrm{d} x =$	$k \int f(x) \, \textrm{d} x$
$\int \big ( f(x) \pm g(x) \big ) \, \textrm{d} x =$	$\int f(x) \, \textrm{d} x \pm \int g(x) \, \textrm{d} x$
$\int x^n \, \textrm{d} x =$	$\frac {1}{n+1} \, x^{n+1} + C$
$\int \sin(x) \, \textrm{d}x =$	$-\cos(x) + C$
$\int \cos(x) \, \textrm{d}x =$	$\sin(x) + C$
$\int \sec^2(x) \, \textrm{d}x =$	$\tan(x) + C$
$\int \csc^2(x) \, \textrm{d}x =$	$-\cot(x) + C$
$\int \sec(x)\tan(x) \, \textrm{d}x =$	$\sec(x) + C$
$\int \csc(x)\cot(x) \, \textrm{d}x =$	$-\csc(x) + C$
$\sum_{i=1}^n c =$	$n \cdot c$
$\sum_{i=1}^n i =$	$\frac{n(n+1)}{2}$
$\sum_{i=1}^n i^2 =$	$\frac{n(n+1)(2n+1)}{6}$
$\int \frac{1}{1+x^2}\,dx=\tan^{-1}x+C$	$\tan^{-1}x+C$

Resource summary

	Created by Bill Andersen about 9 years ago