JEE Main 2021 — Definite Integration Question with Solution

$$\sqrt{{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2}$$

$$=\sqrt{{\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)}^2}$$

$$=\left|\frac{x+1}{x-1}-\frac{x-1}{x+1}\right|$$

$$=\left|\frac{{(x+1)}^2-{(x-1)}^2}{x^2-1}\right|$$

$$=\left|\frac{x^2+2x+1-x^2+2x-1}{x^2-1}\right|$$

$$=\left|\frac{4x}{x^2-1}\right|$$

$$\therefore$$ $$I=\underset{-\frac{1}{\sqrt{2}}}{\overset{\frac{1}{\sqrt{2}}}{\int }}{\left[{\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2\right]}^{\frac{1}{2}}dx$$

$$I=\underset{\frac{1}{\sqrt{2}}}{\overset{\frac{1}{\sqrt{2}}}{\int }}\left|\frac{4x}{x^2-1}\right|dx$$

Let $$f(x)=\left|\frac{4x}{x^2-1}\right|$$

$$\therefore$$ $$f(-x)=\left|\frac{4(-x)}{{(-x)}^2-1}\right|=\left|\frac{-4x}{x^2-1}\right|=\left|\frac{4x}{x^2-1}\right|$$

$$\Rightarrow f(x)=f(-x)$$

$$\therefore$$ f(x) is a even function.

$$\therefore$$ $$I=2{\int }_0^{\frac{1}{\sqrt{2}}}\left|\frac{4x}{x^2-1}\right|dx$$

Using property, If f(x) is an even function then,

$${\int }_{-a}^{a}f(x)=2{\int }_0^{a}f(x)dx$$

x > 0 when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\Rightarrow$$ x² > 0 when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\Rightarrow$$ x² $$-$$ 1 < 0 when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\Rightarrow \frac{1}{x^2-1}<0$$ when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\Rightarrow \frac{4x}{x^2-1}<0$$ when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\therefore$$ $$\left|\frac{4x}{x^2-1}\right|=-\frac{4x}{x^2-1}$$ when $$x\in \left[0,\frac{1}{\sqrt{2}}\right]$$

$$\therefore$$ $$I=2{\int }_0^{\frac{1}{\sqrt{2}}}-\left(\frac{4x}{x^2-1}\right)dx$$

$$=-4{\int }_0^{\frac{1}{\sqrt{2}}}\frac{2x}{x^2-1}dx$$

$$=-4{\left[{\log }_e\left|x^2-1\right|\right]}_0^{\frac{1}{\sqrt{2}}}$$

$$=-4\left[\log \left|\frac{1}{2}-1\right|-\log \left|-1\right|\right]$$

$$=-4{\log }_e\left|-\frac{1}{2}\right|$$

$$=-4{\log }_e\frac{1}{2}$$

$$=-4{\log }_e^{2^{-1}}$$

$$=4{\log }_e^2$$

$$={\log }_e^{2^4}$$

$$={\log }_e^{16}$$