Question

How to integrate
3x^2/(1+x^3)

Answer 1

Topic :-

Indefinite Integration

To Solve :-

$\displaystyle \int\dfrac{3x^2}{1+x^3}\:dx$

Solution :-

$\displaystyle \int\dfrac{3x^2}{1+x^3}\:dx$

$Substitute\:t=1+x^3,$

$dt=3x^2\:dx$

$\displaystyle \int\dfrac{1}{t}\:dt$

$\ln t+C$

$where\:C\:is\:constant\:of\:integration.$

$\left(\because \displaystyle \int\dfrac{1}{z}\:dz=\ln z+C\right)$

$Put\:back\:value\:of\:'t',$

$\ln (1+x^3)+C$

Answer :-

$\underline{\boxed{\displaystyle \int\dfrac{3x^2}{1+x^3}\:dx=ln(1+x^3)+C}}$

Additional Formulae :-

$\displaystyle\int dx=x+C$

$\displaystyle\int x^n\:dx=\dfrac{x^{n+1}}{n+1}+C$

$\displaystyle\int e^x\:dx=e^x+C$

$\displaystyle\int a^x\:dx=\dfrac{a^x}{\ln a}+C$

$\displaystyle\int\sin x\:dx=-\cos x+C$

$\displaystyle\int\cos x\:dx=\sin x+C$

$\displaystyle\int\tan x\:dx=\ln |\sec x|+C$

$\displaystyle\int\csc x\:dx=\ln|\csc x-\cot x| +C$

$\displaystyle\int\sec x\:dx=\ln|\sec x+\tan x| +C$

$\displaystyle\int\cot x\:dx=\ln |\sin x|+C$

$\displaystyle\int\sec^2 x\:dx=\tan x+C$

$\displaystyle\int\csc^2 x\:dx=-\cot x+C$

$\displaystyle\int\sec x\cdot \tan x\:dx=\sec x+C$

$\displaystyle\int\csc x\cdot \cot x\:dx=-\csc x+C$