∫dxxx12−1 =∫x5x6x12-1dx Put x6=t 6x5dx=dt =16∫dttt2-1 x5dx=16dt =16sec-1t+c =16sec-1x6+c =16π2-cosec-1x6 +c =-16cosec-1x6+k =-16sin-1x-6+k

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$\int \frac{d x}{x \sqrt{x^{12}} - 1} =$

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$\frac{- 1}{6} \cos^{- 1} (- x^{- 6}) + c$

$\frac{- 1}{6} \sin^{- 1} (+ x^{- 6}) + c$

$\frac{- 1}{6} \cos^{- 1} (x^{6}) + c$

$\frac{1}{6} \sin^{- 1} (x^{6}) + c$

answer is B.

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Detailed Solution

$\int \frac{dx}{x \sqrt{x^{12}} - 1} = \int \frac{x^{5}}{x^{6} \sqrt{x^{12} - 1}} dx Put x^{6} = t 6 x^{5} dx = dt = \frac{1}{6} \int \frac{dt}{t \sqrt{t^{2} - 1}} x^{5} dx = \frac{1}{6} dt = \frac{1}{6} \sec^{- 1} t + c = \frac{1}{6} \sec^{- 1} (x^{6}) + c = \frac{1}{6} (\frac{π}{2} - {cosec}^{- 1} x^{6}) + c = \frac{- 1}{6} {cosec}^{- 1} x^{6} + k = \frac{- 1}{6} \sin^{- 1} (x^{- 6}) + k$