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Mathematics Question on Some Properties of Definite Integrals

The value of the integral ∫−1/21/2[(x+1x−1)2+(x+1x−1)2−2]1/2  dx\int\limits ^{1/2}_{-1/2}\left[\left(\frac{x+1}{x-1}\right)^{^2}+\left(\frac{x+1}{x-1}\right)^{^2}-2\right]^{^{1/2}}\:\:dx is

A

log⁡(43)\log\left(\frac{4}{3}\right)

B

4 log⁡(34)4\,\log\left(\frac{3}{4}\right)

C

4 log⁡(43)4\,\log\left(\frac{4}{3}\right)

D

log⁡(34)\log\left(\frac{3}{4}\right)

Answer

4 log⁡(43)4\,\log\left(\frac{4}{3}\right)

Explanation

Solution

∫−1/21/2[(x+1x−1)2+(x−1x+1)2−2]1/2dx\int\limits_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right]^{1 / 2} d x
=∫−1/21/2[(x+1x−1−x−1x+1)2]1/2dx=\int\limits_{-1 / 2}^{1 / 2}\left[\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)^{2}\right]^{1 / 2} d x
=∫−1/21/2∣4xx2−1∣dx=\int\limits_{-1 / 2}^{1 / 2}\left|\frac{4 x}{x^{2}-1}\right| d x
=∫−1/20∣4x1−x2∣dx+∫01/2∣4x1−x2∣dx=\int\limits_{-1 / 2}^{0}\left|\frac{4 x}{1-x^{2}}\right| d x+\int\limits_{0}^{1 / 2}\left|\frac{4 x}{1-x^{2}}\right| d x
=−4∫−1/20x1−x2dx+4∫01/2x1−x2dx=-4 \int\limits_{-1 / 2}^{0} \frac{x}{1-x^{2}} d x+4 \int_{0}^{1 / 2} \frac{x}{1-x^{2}} d x
=2\left\\{\log \left(1-x^{2}\right\\}_{-1 / 2}^{0}-2\left\\{\log \left(1-x^{2}\right)\right\\}_{0}^{1 / 2}\right.
=−2log⁡(1−14)−2log⁡(1−14)=-2 \log \left(1-\frac{1}{4}\right)-2 \log \left(1-\frac{1}{4}\right)
=−4log⁡34=4log⁡43=-4 \log \frac{3}{4}=4 \log \frac{4}{3}