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Q.

$\int \frac{\sqrt{x - 1}}{x \sqrt{x + 1}} d x is equal to$

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a

$\ln |x - \sqrt{x^{2} - 1}| - \tan^{- 1} x + C$

b

$\ln |x + \sqrt{x^{2} - 1}| - \tan^{- 1} x + C$

c

$\ln |x + \sqrt{x^{2} - 1}| - \sec^{- 1} x + C$

d

$\ln |x - \sqrt{x^{2} - 1}| - \sec^{- 1} x + C$

answer is D.

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Detailed Solution

$\begin{aligned} I = \int \frac{\sqrt{x - 1}}{x \sqrt{x + 1}} dx = \int \frac{x - 1}{x \sqrt{x^{2} - 1}} \\ = \int \frac{dx}{\sqrt{x^{2} - 1}} - \int \frac{dx}{x \sqrt{x^{2} - 1}} \\ = \ln |x + \sqrt{x^{2} - 1}| - \sec^{- 1} x + C \end{aligned}$

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