Question

$\int e^{{\tan }^{-1}x}(1+x+x^2)d\left({\cot }^{-1}x\right)=?$

Answer 1

Consider the given integral.

$I=\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx$

Let $t={\tan }^{-1}x$

$dt=\frac{dx}{1+x^2}$

Therefore,

$I=\int e^{t}dt$

$I=e^t+C$

On putting the value of $t$, we get

$I=e^{{\tan }^{-1}x}+C$

Hence, this is the answer.