Question

Evaluate :- $\int e^{{\tan }^{-1}x}.\frac{1+x+x^2}{1+x^2}dx.$

Answer 1

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

Let ${\tan }^{-1}x=t\Rightarrow x=\tan t,x^2={\tan }^{2}t\frac{1}{1+x^2}dx=dtI=\int e^t\frac{1+\tan t+{\tan }^{2}t}{1+x^2}(1+x^2)dt=\int e^t(1+\tan t+{\tan }^{2}t)dt=\int e^t(\tan t+{\sec }^{2}t)dt\{\int e^x(f\left(x\right)+f^{{}^{\prime }}\left(x\right))=e^{x}f\left(x\right)+C\}=e^t\left(\tan t\right)+C=x{e}^{{\tan }^{-1}x}+C$