JEE Main 2025 Apr 4 Shift 1, Mathematics Q16: Some Properties of Definite Integrals

$I={\int }_{-1}^1\frac{\left(1+\sqrt{|x|-x}\right)e^x+\left(\sqrt{|x|-x}\right)e^{-x}}{e^x+e^{-x}}dx$
Apply the king's rule ${\int }_{-a}^{a}f(x)dx={\int }_{-a}^{a}f(-x)dx$: replacing $x$ by $-x$ swaps $|x|-x$ with $|x|+x$ and swaps $e^x\leftrightarrow e^{-x}$.

$I=\displaystyle\int_{-1}^{1} \frac{\left(1+\sqrt{|-x|-(-x)}\right)e^{-x} +\left(\sqrt{|-x|-(-x)}\right)e^{-(-x)}} {e^{-x}+e^{-(-x)}}\,dx$

$\Rightarrow I=\displaystyle\int_{-1}^{1} \frac{\left(1+\sqrt{|x|+x}\right)e^{-x} +\left(\sqrt{|x|+x}\right)e^{x}} {e^{-x}+e^{x}}\,dx$

Adding both forms:

$\Rightarrow 2I=\displaystyle\int_{-1}^{1} \frac{\left(1+\sqrt{|x|+x}+\sqrt{|x|-x}\right)(e^{x}+e^{-x})} {e^{-x}+e^{x}}\,dx$

$\Rightarrow 2I=\displaystyle\int_{-1}^{1} \left(1+\sqrt{|x|+x}+\sqrt{|x|-x}\right)\,dx$

$\Rightarrow 2I=2\displaystyle\int_{0}^{1} \left(1+\sqrt{|x|+x}+\sqrt{|x|-x}\right)\,dx$

$\Rightarrow 2I=2\displaystyle\int_{0}^{1} \left(1+\sqrt{2x}+\sqrt{0}\right)\,dx$

Note: $|x|=x$ in the interval [0, 1]

$\Rightarrow I=\displaystyle\int_{0}^{1}(1+\sqrt{2x})\,dx$

$={\left[x+\frac{2\sqrt{2}}{3}{x}^{3/2}\right]}_0^1$

$\therefore \boxed{I=1+\frac{2\sqrt{2}}{3}}$