$\int \frac{1}{x^{2} (x^{4} + 1)^{3 / 4}} d x$ is equal to ____

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

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

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

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

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Solution

The correct option is C $\frac{- (1 + x^{4})^{1 / 4}}{x} + C$
Let $I = \int \frac{1}{x^{2} (x^{4} + 1)^{3 / 4}} d x = \int \frac{1}{x^{2} [x^{4} (1 + \frac{1}{x^{4}})]^{3 / 4}} d x$
$= \int \frac{1}{x^{2} \cdot x^{3} (1 + \frac{1}{x^{4}})^{3 / 4}} d x = \int \frac{1}{x^{5} (1 + \frac{1}{x^{4}})^{3 / 4}}$

Substitute $1 + \frac{1}{x^{4}} = u$ $\Rightarrow - \frac{4}{x^{5}} d x = d u$

Therefore $I = \int \frac{1}{4 u^{3 / 4}} d u = - \frac{1}{4} \int u^{- 3 / 4} d u$
$= - u^{1 / 4} + C = {(1 + \frac{1}{x^{4}})}^{1 / 4} + C$
$= \frac{- (1 + x^{4})^{1 / 4}}{x} + C$

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Similar questions

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

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

The intergral $\int \frac{d x}{x^{2} (x^{4} + 1)^{3 / 4}}$ eauals:

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

m

[\begin{matrix} 2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & - 4 \end{matrix}] [\begin{matrix} x \\ 4 \\ - 1 \end{matrix}]

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