Question

${\int }_{-1}^1\frac{1}{1+x^2+e^x+x^2{e}^x}dx$

Answer 1

Given,

$\underset{-1}{\overset{1}{\int }}\frac{1}{1+x^2+e^x+x^2{e}^x}dx$

Now,

$\begin{array}{c}I={\int }_{-1}^1\frac{dx}{1+x^2+e^x+x^2{e}^x}\\ I={\int }_{-1}^1\frac{dx}{\left(1+x^2\right)\left(1+e^x\right)}\\ I={\int }_{-1}^1\frac{e^{x}dx}{\left(1+x^2\right)\left(e^x+1\right)}\\ 2I={\int }_{-1}^1\frac{dx}{1+x^2}\\ I={\int }_0^1\frac{dx}{1+x^2}\\ I={\left[{\tan }^{-1}x\right]}_0^1=\frac{\pi }{4}\end{array}$