The value of int frac{d x}{x^2(x^4+1)^{frac{3 | vedclass

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(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$.We can rewrite the integral as $I = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{\frac{3}{

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$-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+c$,where $c$ is a constant of integration.

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(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$.We can rewrite the integral as $I = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{\frac{3}{4}}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{\frac{3}{4}}}$.Let $t = 1 + \frac{1}{x^4}$. Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.Substituting these into the integral,we get:$I = -\frac{1}{4} \int t^{-\frac{3}{4}} dt$.Integrating with respect to $t$,we have:$I = -\frac{1}{4} \cdot \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + c = -t^{\frac{1}{4}} + c$.Substituting back $t = 1 + \frac{1}{x^4}$,we get:$I = -(1 + \frac{1}{x^4})^{\frac{1}{4}} + c = -\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}} + c$.

The value of $\int \frac{d x}{x^2(x^4+1)^{\frac{3}{4}}}$ is

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