Practicing Success
Practicing Success
CUET Preparation Today
$\int\frac{e^{\tan^{-1}}x}{(1+x^2)}(1+x+x^2)dx$ is equal to : |
$\frac{e^{\tan^{-1}x}}{1+x^2}$ $e^{\tan^{-1}x}.(1+x^2)$ $xe^{\tan^{-1}x}$ None of these |
$xe^{\tan^{-1}x}$ |
Put $\tan^{-1}x=z$ i.e. x = tan z $∴\int\frac{e^{\tan^{-1}}x}{1+x^2}(1+x+x^2)dx=\int e^z(1+\tan z+\tan^2z)dz=\int e^z(\tan z+sec^2z)dz$ $=\int e^z(\tan z+\frac{d}{dz}(\tan z))dz=e^z\tan z+C=xe^{\tan^{-1}x}+C$ |
