Question

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

Answer 1

Let $I=\int \frac{x{e}^x}{{(x+1)}^2}dx$

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

$=\int e^x[{\frac{(x+1)}{(x+1)}}^2-\frac{1}{{(x+1)}^2}]dx$

$=\int e^x[\frac{1}{(x+1)}-\frac{1}{{(x+1)}^2}]dx$

Since, $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+C$

$\therefore \frac{x{e}^x}{{(x+1)}^2}dx=e^x\left[\frac{1}{(x+1)}\right]+C$