☞ 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 vdx ]dx + c
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