$\int \frac{1}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}} d x =$

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

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

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

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

answer is C.

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Detailed Solution

$I = \int \frac{d x}{x^{2} {(x^{4} + 1)}^{\frac{3}{4}}}$

$\begin{array}{l} = \int \frac{d x}{x^{2} {[x^{4} (1 + \frac{1}{x^{4}})]}^{\frac{3}{4}}} \\ = \int \frac{d x}{x^{2} - x^{3} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}} \\ = \int \frac{d x}{x^{5} {(1 + \frac{1}{x^{4}})}^{\frac{3}{4}}} \\ Put 1 + \frac{1}{x^{4}} = t \\ Differentiating both sides \end{array}$

$\begin{aligned} \frac{- 4}{x^{5}} d x = d t \\ \frac{d x}{x^{5}} = \frac{- d t}{4} \end{aligned}$

$\begin{array}{l} I = \frac{- 1}{4} \int \frac{d t}{t^{\frac{3}{4}}} \\ = \frac{- 1}{4} (\frac{t^{\frac{- 3}{4}} + 1}{- 3} + 1) + c \\ = \frac{- 1}{4} (\frac{t^{\frac{1}{4}}}{\frac{1}{4}}) + c \\ = - t^{\frac{1}{4}} + c \\ = - {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} + C \end{array}$