$\int \frac{\sin^{- 1} x - \cos^{- 1} x}{\sin^{- 1} x + \cos^{- 1} x} d x =$

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$\frac{4}{π} [x \sin^{- 1} x + \sqrt{1 - x^{2}}] - x + c$

$\frac{2}{π} [x \sin^{- 1} x + \sqrt{1 - x^{2}}] - x + c$

$\frac{1}{π} [x \sin^{- 1} x + \sqrt{1 - x^{2}}] - x + c$

$\frac{2}{π} [x \sin^{- 1} x + \sqrt{1 + x^{2}}] - x + c$

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Detailed Solution

$\int \frac{\sin^{- 1} x - \cos^{- 1} x}{\sin^{- 1} x + \cos^{- 1} x} d x$

$= \int \frac{\sin^{- 1} x - (\frac{π}{2} - \sin^{- 1} x)}{π / 2} d x (∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2})$

$= \frac{2}{π} \int (2 \sin^{- 1} x - \frac{π}{2}) d x = \frac{4}{π} \int \sin^{- 1} x 1. d x - \int 1 d x$

$= \frac{4}{π} [\sin^{- 1} x . x - \int \frac{1}{\sqrt{1 - x^{2}}} x d x] - x + c$

$= \frac{4}{π} [x \sin^{- 1} x + \frac{1}{2} \int \frac{- 2 x}{\sqrt{1 - x^{2}}} d x] - x + c$

$= \frac{4}{π} [x \sin^{- 1} x + \frac{1}{2} 2 \sqrt{1 - x^{2}}] - x + c$

$= \frac{4}{π} [x \sin^{- 1} x + \sqrt{1 - x^{2}}] - x + c$

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