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

Question

The value of $\int_\limits{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x$ is equal to

Accepted Answer

$$\begin{aligned}
& I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \
& =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}}
\end{aligned}$$

$$\begin{aligned}
& +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \
= & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \
= & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x
\end{aligned}$$

$$\begin{aligned}
& =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \
& =1+\frac{2 \sqrt{2}}{3}
\end{aligned}$$

Solutions · 1