The value of $\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:

(D) $4 \log _{e}(3+2 \sqrt{2})$

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Definite Integrals

The value of

\int_{\frac{- 1}{2}}^{\frac{1}{2}} ((\frac{x + 1}{x - 1})^{2} + (\frac{x - 1}{x + 1})^{2} - 2)^{1/2} d x

is:

Answer: (A)

With

u = \frac{x + 1}{x - 1}

the expression under the root is

u^{2} + \frac{1}{u ^{2}} - 2 = (u - \frac{1}{u})^{2}

, so the integrand is

u - \frac{1}{u}

, and

u - \frac{1}{u} = \frac{( x + 1 ) ^{2} - ( x - 1 ) ^{2}}{x ^{2} - 1} = \frac{4 x}{x ^{2} - 1} .

[- \frac{1}{2}, \frac{1}{2}]

one has

x^{2} < 1

, so the integrand equals

\frac{4 ∣ x ∣}{1 - x ^{2}}

, which is even. Hence

\int_{- 1/ 2}^{1/ 2} \frac{4 ∣ x ∣}{1 - x ^{2}} d x = 8 \int_{0}^{1/ 2} \frac{x}{1 - x ^{2}} d x = - 4 [lo g_{e} (1 - x^{2})]_{0}^{1/ 2} = - 4 lo g_{e} \frac{1}{2} = lo g_{e} 16.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option