JEE Main 2025 — 4 April, Morning Shift — Question 32

$I=\int_{-1}^{1} \frac{(1+\sqrt{|x|-x}) e^{x}+(\sqrt{|x|-x}) e^{-x}}{e^{x}+e^{-x}} d x$

$$\begin{aligned} & =\int_{0}^{1}\left(\frac{(1+\sqrt{|x|-x}) e^{x}+(\sqrt{|x|-x}) e^{-x}}{e^{x}+e^{-x}}\right. \\& \quad+\left(\frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{-x}}{e^{-x}+e^{x}}\right) d x \end{aligned}$$ $$=\int_{0}^{1} \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})\left(e^{x}+e^{-x}\right)}{e^{x}+e^{-x}} d x$$ $$=\int_{0}^{1}(1+\sqrt{|x|-x}+\sqrt{|x|+x}) d x$$

$=\int_{0}^{1}(1+\sqrt{2 x}) d x=\left.x\right|_{0} ^{1}+\left.\frac{\sqrt{2} x^{\frac{3}{2}}}{\frac{3}{2}}\right|_{0} ^{1}$

$=1+\frac{2 \sqrt{2}}{3}$