JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 9)
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
$1+\frac{2 \sqrt{2}}{3}$
$1-\frac{2 \sqrt{2}}{3}$
$2+\frac{2 \sqrt{2}}{3}$
$3-\frac{2 \sqrt{2}}{3}$
Explanation
$$\begin{aligned} & I=\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x \\ & =\int_0^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} \end{aligned}$$
$$\begin{aligned} & +\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^x}\right) d x \\ = & \int_0^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^x+e^{-x}\right)}{e^x+e^{-x}} d x \\ = & \int_0^1(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x \end{aligned}$$
$$\begin{aligned} & =\int_0^1(1+\sqrt{2 x}) d x=\left.x\right|_0 ^1+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_0 ^1 \\ & =1+\frac{2 \sqrt{2}}{3} \end{aligned}$$
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