int frac{dx}{x^2(x^4+1)^{3/4}} equals | vedclass

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(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.Take $x^4$ common from the bracket: $I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} (1 + \frac{1}{x^4})^{3/4}

Vedclass · Accepted Answer

$-\left(\frac{x^4+1}{x^4}\right)^{1/4}+c$

Vedclass · Answer

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.Take $x^4$ common from the bracket: $I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.Let $t = 1 + \frac{1}{x^4} = 1 + x^{-4}$.Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.Substituting these into the integral:$I = \int (1 + x^{-4})^{-3/4} (x^{-5} dx) = \int t^{-3/4} \left(-\frac{1}{4} dt\right) = -\frac{1}{4} \int t^{-3/4} dt$.Integrating with respect to $t$:$I = -\frac{1}{4} \left(\frac{t^{1/4}}{1/4}\right) + c = -t^{1/4} + c$.Substituting back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:$I = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c$.

$\int \frac{dx}{x^2(x^4+1)^{3/4}}$ equals

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