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

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(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.Factor out $x^4$ from the parenthesis: $I = \int \frac{dx}{x^2(x^4(1 + x^{-4}))^{3/4}} = \int \frac{dx}

Vedclass · Accepted Answer

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

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(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.Factor out $x^4$ from the parenthesis: $I = \int \frac{dx}{x^2(x^4(1 + x^{-4}))^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1 + x^{-4})^{3/4}} = \int \frac{dx}{x^5(1 + x^{-4})^{3/4}}$.Let $t = 1 + x^{-4}$.Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.Substituting these into the integral:$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.Substituting back $t = 1 + x^{-4} = \frac{x^4+1}{x^4}$:$I = -(\frac{x^4+1}{x^4})^{1/4} + c$.

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

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