How do I evaluate $\int_{-1}^1\frac{dx}{(1+x^2)(e^x+1)}$?

Guys I got the answer $$\int_{-1}^1\frac{dx}{(1+x^2)(e^{-x}+1)}$$ the numerator and denominator were multiplied with $e^x$ so the term $(e^{-x}+1)$ changed to $$e^{-x+x} \Rightarrow e^{0} = 1$$ and $1$ in $(e^{-x}+1)$ changed to $e^{x}$, so finally the denominator is equal to $$1+e^x \Leftrightarrow e^x+1.$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ There is an identity $\ds{\pars{~\Theta\ \mbox{is the}\ Heaviside\ Theta\ Function~}}$: \begin{align} &{1 \over \expo{x} + 1} = \left\{\begin{array}{lcl} \ds{\underbrace{\Theta\pars{-x} + {\on{sgn}\pars{x} \over \expo{\verts{x}} + 1}} _{\ds{\substack{\mbox{Note that the}\ } \\ second\ term\ \mbox{is an} \\ \underline{odd\ function}.}}} & \mbox{if} & \ds{x \not= 0} \\[3mm] \ds{1 \over 2} & \mbox{if} & \ds{x = 0} \end{array}\right. \\[1cm] & \mbox{Then,}\quad \int_{-1}^{1}{\dd x \over \pars{1 + x^{2}}\pars{\expo{x} + 1}} \\[2mm] & \phantom{\mbox{Then,}\ }= \int_{-1}^{0}{\dd x \over 1 + x^{2}} = \bbx{\pi \over 4} \\ & \end{align}