The value of _-1^1(1+|x|-x)e^x+(|x|-x)e^-xe^x+e^-xdx is equal to

Question

The value of $\int _{-1}^{1}\frac{(1+\sqrt{|x|-x}){e}^{x}+(\sqrt{|x|-x}){e}^{-x}}{{e}^{x}+{e}^{-x}}dx$ is equal to

Accepted Answer

$\begin{gathered} I={\int }_{1}^{-1}\frac{(1+\sqrt{|x|-x}){e}^{x}+(\sqrt{|x|-x}){e}^{-x}}{{e}^{x}+{e}^{-x}}dx \ ={\int }_{1}^{0}\frac{(1+\sqrt{|x|-x}){e}^{x}+(\sqrt{|x|-x}){e}^{-x}}{{e}^{x}+{e}^{-x}} \end{gathered}$

$\begin{gathered} +(\frac{(1+\sqrt{|x|+x}){e}^{-x}+(\sqrt{|x|+x}){e}^{-x}}{{e}^{-x}+{e}^{x}})dx \ = {\int }_{1}^{0}\frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})({e}^{x}+{e}^{-x})}{{e}^{x}+{e}^{-x}}dx \ = {\int }_{1}^{0}(1+\sqrt{|x|-x}+\sqrt{|x|+x})dx \end{gathered}$

$\begin{gathered} ={\int }_{1}^{0}(1+\sqrt{2x})dx={x|}_{1}^{0}+{\frac{\sqrt{2}{x}^{\frac{3}{2}}}{\frac{3}{2}}|}_{1}^{0} \ =1+\frac{2\sqrt{2}}{3} \end{gathered}$

Solutions · 1