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$\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x$ is equal to |
$x e^{\tan ^{-1} x}+c$ $x^2 e^{\tan ^{-1} x}+c$ $\frac{1}{x e^{\tan ^{-1} x}}+c$ $\frac{1}{x^2} e^{\tan ^{-1} x}+c$ |
$x e^{\tan ^{-1} x}+c$ |
$I=\int e^p\left(\sec ^2 p+\tan p\right) d p$ (If $p=\tan ^{-1} x \Rightarrow x=\tan p \Rightarrow d x=\sec ^2 p d p$) $=e^{p} \tan p=x e^{\tan ^{-1} x}+c$ Hence (1) is the correct answer. |
