∫dxx4x4+14=∫dxx4x41+1x41/4=∫dxx5⋅1+1x41/4 Put 1+1x4=t,−4x5dx=dt=−14∫dtt1/4=−14t3/443+c=−131+1x43/4+c

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$\int \frac{1}{x^{4} \sqrt[4]{x^{4} + 1}} d x =$

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$\frac{1}{3} {(1 + \frac{1}{x^{4}})}^{3 / 4} + C$

$\frac{- 1}{3} {(1 + \frac{1}{x^{4}})}^{3 / 4} + C$

$\frac{2}{3} {(1 + \frac{1}{x^{4}})}^{3 / 4} + C$

$\frac{- 2}{3} {(1 + \frac{1}{x^{4}})}^{3 / 4} + C$

answer is B.

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Detailed Solution

$\begin{array}{l} \int \frac{dx}{x^{4} \sqrt[4]{x^{4} + 1}} = \int \frac{dx}{x^{4} {[x^{4} (1 + \frac{1}{x^{4}})]}^{1 / 4}} \\ = \int \frac{dx}{x^{5} \cdot {(1 + \frac{1}{x^{4}})}^{1 / 4}} Put 1 + \frac{1}{x^{4}} = t, \frac{- 4}{x^{5}} dx = dt \\ = \frac{- 1}{4} \int \frac{dt}{t^{1 / 4}} \\ = (- \frac{1}{4}) t^{3 / 4} \frac{4}{3} + c \\ = \frac{- 1}{3} {(1 + \frac{1}{x^{4}})}^{3 / 4} + c \end{array}$