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

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Question

(A) Let $I = \int \frac{dx}{x^2(1+x^4)^{3/4}}$.We can rewrite the integral by taking $x^4$ common from the bracket:$I = \int \frac{dx}{x^2 \cdot (x^4(

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x^2(1+x^4)^{3/4}}$.We can rewrite the integral by taking $x^4$ common from the bracket:$I = \int \frac{dx}{x^2 \cdot (x^4(x^{-4}+1))^{3/4}} = \int \frac{dx}{x^2 \cdot x^3(1+x^{-4})^{3/4}} = \int \frac{dx}{x^5(1+x^{-4})^{3/4}}$.Let $t = 1+x^{-4}$.Then $dt = -4x^{-5} dx$,which implies $x^{-5} dx = -\frac{1}{4} dt$.Substituting these into the integral:$I = \int \frac{-\frac{1}{4} dt}{t^{3/4}} = -\frac{1}{4} \int t^{-3/4} dt$.Integrating with respect to $t$:$I = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + C = -t^{1/4} + C$.Substituting $t = 1+x^{-4}$ back:$I = -(1+x^{-4})^{1/4} + C = -(\frac{x^4+1}{x^4})^{1/4} + C = -\frac{(1+x^4)^{1/4}}{x} + C$.

$\int {\frac{{dx}}{{{x^2}{{(1 + {x^4})}^{\frac{3}{4}}}}}} $ is equal to- (where $C$ is constant of integration)

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