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  4. ∫ e tan -1 x(1+x+x2) ⋅ d( cot -1 x) is equal to

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Q. $\int e^{\tan ^{-1} x}\left(1+x+x^{2}\right) \cdot d\left(\cot ^{-1} x\right)$ is equal to

Integrals

Solution:

$I =\int e^{\tan ^{-1} x}\left(1+x+x^{2}\right) \cdot\left(-\frac{1}{1+x^{2}}\right) d x$
$=-\int e^{\tan ^{-1} x}\left(1+\frac{x}{1+x^{2}}\right) d x$
$=-\int e^{\tan ^{-1} x} d x-\int x \cdot \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x$
$=-\int e^{\tan ^{-1} x} d x-x \cdot e^{\tan ^{-1} x}+\int e^{\tan ^{-1} x} d x+ C$
$=-x \cdot e^{\tan ^{-1} x}+C$