Formula of Integral Calculus | Basic Integration Formulas | ইন্টিগ্রেশন ক্যালকুলাস এর সূত্রসমূহ

✪➤ Formula of Integral Calculus | Basic Integration Formulas
☞ 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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Formulas of Differential Calculus | Derivative Rules | ক্যালকুলাস এর সূত্রসমূহ

Formulas of Differential Calculus | Derivative Rules | ক্যালকুলাস এর সূত্রসমূহ

$\frac{d}{dx}(c)$  = 0                             [where, c is single constant]

$\frac{d}{dx}(cu)$ = c $\frac {d}{dx}$ (u)   [where, c is with a variable]

$\frac{d}{dx}$ (xⁿ) = nx^n-1

$\frac{d}{dx}$ $(\sqrt {x})= \frac {1}{2\sqrt{x}}$

$\frac{d}{dx}$ (e^x)= e^x

$\frac{d}{dx}$ (e^mx)= me^mx

$\frac{d}{dx}$ (e^-mx)= -me^-mx

$\frac{d}{dx}$ (log_a x)= $\frac {1}{x}$ log_ae

$\frac{d}{dx}$ (log x)= $\frac {1}{x}$

$\frac{d}{dx}$ (sin x)= cos x

$\frac{d}{dx}$ (sin mx)= m cos mx

$\frac{d}{dx}$ (cos x)= - sin x

$\frac{d}{dx}$ (cos mx)= - m sin mx

$\frac{d}{dx}$ (sec x)= sec x. tan x

$\frac{d}{dx}$ (cosec x)= -cosec x. cot x

$\frac{d}{dx}$ (tan x)= sec² x

$\frac{d}{dx}3$ (cot x)= -cosec² x

$\frac{d}{dx}$ (sin^-1) = $1  \over { \sqrt{1-x^2 }}$

$\frac{d}{dx}$ (cos^-1)= $-1  \over { \sqrt{1-x^2 }}$

$\frac{d}{dx}$ (tan^-1)= $\frac{1}{1+x^2}$

$\frac{d}{dx}$ (uv)= u $ \frac {d}{dx}$ (v) + v $\frac {d}{dx}$ (u)

$\frac{d}{dx}$ (uvw) = vw $\frac {d}{dx}$ (u) + uw $\frac {d}{dx}$ (v) + uv $\frac {d}{dx}$( w )

$\frac {d}{dx}$ ($\frac {u}{v})$ = $\frac{v\frac{\mathrm{d}}{\mathrm{d} x} ( u )-u\frac{\mathrm{d} }{\mathrm{d} x} ( v )}{v^{2}}$


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