Question

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

Answer 1

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

Let t = ${\tan }^{-1}x$
$x=\tan t$
$\frac{1}{1+x^2}dx=dt$

$\begin{array}{c}I=\int e^t\left(1+\mathrm{tant}+{\tan }^{2}t\right)dt\\ I=\int e^t\left(\mathrm{tant}+{\sec }^{2}t\right)dt\end{array}$

We know that,

$\begin{array}{c}\int e^x\left[f\left(x\right)+f^{\prime }\left(x\right)\right]dx\\ =e^{x}f\left(x\right)+C\end{array}$
$\begin{array}{c}=e^t\tan t+C\\ =x{e}^{{\tan }^{-1}x}+C\end{array}$