☞ The remark that integration is (almost) an inverse to the operation of differentiation means that if
$\frac {d}{dx}f(x) = g(x)$ [Differentiation]
then, $\int g(x) dx = f(x) + c$ [Integration]
✪Most Used Formula:
$\int dx= x+ c$
$\int x dx = \frac{x^2}{2}+c$
$\int x^n dx = \frac{x^n+1}{n+1}+c$
$\int e^x dx = e^x+c$
$\int e^{mx} dx$ = $ \Large \frac {e^{mx}}{m}+c$
$\int e^{-mx} dx$ = $\Large \frac {e^{-mx}}{-m}+c$
$\int \frac{1}{x} dx= log x+c$
$\int \sin x\; dx= -\cos x\;+c$
$\int \sin mx\; dx$= $\Large \frac {-cos mx}{m}+c$
$\int \cos x\; dx= \sin x\;+c$
$\int \cos mx\; dx$= $\Large \frac {sin mx}{m}+c$
$\int sec^2 x dx= \tan x\;+c$
$\int cosec^2 x dx= -\cot x\;+c$
$\int \sec x\; \tan x\; dx= \sec x\;+c$
$\int cosecx \cot x\; dx= -cosec x+c$
$\large \int \frac {1}{1+x^2} dx$= $tan^{-1} x+c$
$\large \int \frac {1}{\sqrt{1-x^2}} dx$ =$ sin^{-1} x+c$
$\int \frac {f'(x)}{f(x)} dx= log f(x)+c$
$\int uv$ $dx$= $u \int v dx - \int[ \frac {d}{dx}(u)$ $\int v$dx$ ]$dx + c
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