Question

The value of the integral \(\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x\) is equal to
(A) \(\log _{e}\left(\frac{4}{3}\right)\)
(B) \(4 \log _{\mathrm{e}}\left(\frac{3}{4}\right)\)
(C) \(4 \log _{\mathrm{e}}\left(\frac{4}{3}\right)\)
(D) \(\log _{e}\left(\frac{3}{4}\right)\)

Answer 1

Ans(C)

Hint \(: \int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x=\int_{-y / 2}^{1 / 2}\left|\frac{x+1}{x-1}-\frac{x-1}{x+1}\right| d x\)
$$
=\int_{-1 / 2}^{0} \frac{-4 x}{-(x-1)(x+1)} d x+\int_{0}^{1 / 2} \frac{4 x}{-(x-1)(x+1)} d x \int_{-\frac{1}{2}}^{0} \frac{4 x}{x^{2}-1}-\int_{0}^{\frac{1}{2}} \frac{4 x}{x^{2}-1} d x=4 \ln 4 / 3
$$

Answer 2

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