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
Solution
<p>$$\begin{aligned}
& I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \\
& =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \\
= & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \\
= & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \\
& =1+\frac{2 \sqrt{2}}{3}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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