The value of \int\limits_ - 1 \over \sqrt 2 ^1 \over \sqrt 2 ( ( x + 1 \over x - 1 )^2 + ( x - 1 \over x + 1 )^2 - 2 )^1 \over 2dx is :

Definite Integration PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q31 with Solution

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Correct answer: B

Let $I=\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2}\,dx.$

Set $a=\frac{x+1}{x-1},\qquad b=\frac{x-1}{x+1}.$ Then $ab=1.$ So, $a^2+b^2-2=a^2+b^2-2ab=(a-b)^2.$

Hence the integrand becomes $\sqrt{(a-b)^2}=|a-b|.$ Now,

=\frac{(x+1)^2-(x-1)^2}{x^2-1} =\frac{4x}{x^2-1}.$$ Therefore, $$\sqrt{\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2} =\left|\frac{4x}{x^2-1}\right|.$$ Since on the interval $\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]$, we have $x^2<1$, so $x^2-1<0$. Thus $$\left|\frac{4x}{x^2-1}\right|=\frac{4|x|}{1-x^2}.$$ So, $$I=\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \frac{4|x|}{1-x^2}\,dx.$$ 2. **Use symmetry** The integrand is even, so $$I=2\int_0^{\frac1{\sqrt2}} \frac{4x}{1-x^2}\,dx =8\int_0^{\frac1{\sqrt2}} \frac{x}{1-x^2}\,dx.$$ 3. **Integrate** Let $$u=1-x^2 \implies du=-2x\,dx \implies x\,dx=-\frac12 du.$$ Then $$\int \frac{x}{1-x^2}\,dx=-\frac12\int \frac{du}{u}=-\frac12\ln|u|=-\frac12\ln(1-x^2).$$ Thus, $$I=8\left[-\frac12\ln(1-x^2)\right]_0^{1/\sqrt2} =-4\left[\ln(1-x^2)\right]_0^{1/\sqrt2}.$$ Now, $$1-\left(\frac1{\sqrt2}\right)^2=1-\frac12=\frac12.$$ So, $$I=-4\left(\ln\frac12-\ln 1\right) =-4\ln\frac12=4\ln 2=\ln 16.$$ 4. **Compare with options** $$\ln 16$$ corresponds to **Option B**. 5. **Verification with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

View written solutionFree

Correct answer: B

Let $I=\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2}\,dx.$

Set $a=\frac{x+1}{x-1},\qquad b=\frac{x-1}{x+1}.$ Then $ab=1.$ So, $a^2+b^2-2=a^2+b^2-2ab=(a-b)^2.$

Hence the integrand becomes $\sqrt{(a-b)^2}=|a-b|.$ Now,

=\frac{(x+1)^2-(x-1)^2}{x^2-1} =\frac{4x}{x^2-1}.$$ Therefore, $$\sqrt{\left(\frac{x+1}{x-1}\right)^2+\left(\frac{x-1}{x+1}\right)^2-2} =\left|\frac{4x}{x^2-1}\right|.$$ Since on the interval $\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]$, we have $x^2<1$, so $x^2-1<0$. Thus $$\left|\frac{4x}{x^2-1}\right|=\frac{4|x|}{1-x^2}.$$ So, $$I=\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \frac{4|x|}{1-x^2}\,dx.$$ 2. **Use symmetry** The integrand is even, so $$I=2\int_0^{\frac1{\sqrt2}} \frac{4x}{1-x^2}\,dx =8\int_0^{\frac1{\sqrt2}} \frac{x}{1-x^2}\,dx.$$ 3. **Integrate** Let $$u=1-x^2 \implies du=-2x\,dx \implies x\,dx=-\frac12 du.$$ Then $$\int \frac{x}{1-x^2}\,dx=-\frac12\int \frac{du}{u}=-\frac12\ln|u|=-\frac12\ln(1-x^2).$$ Thus, $$I=8\left[-\frac12\ln(1-x^2)\right]_0^{1/\sqrt2} =-4\left[\ln(1-x^2)\right]_0^{1/\sqrt2}.$$ Now, $$1-\left(\frac1{\sqrt2}\right)^2=1-\frac12=\frac12.$$ So, $$I=-4\left(\ln\frac12-\ln 1\right) =-4\ln\frac12=4\ln 2=\ln 16.$$ 4. **Compare with options** $$\ln 16$$ corresponds to **Option B**. 5. **Verification with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.