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

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(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$. Taking $x^4$ common from the bracket,we get: $I = \int \frac{1}{x^2 [x^4(1 + \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We have the integral $I = \int \frac{1}{x^2(x^4 + 1)^{3/4}} dx$. Taking $x^4$ common from the bracket,we get: $I = \int \frac{1}{x^2 [x^4(1 + \frac{1}{x^4})]^{3/4}} dx = \int \frac{1}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} dx = \int \frac{1}{x^5 (1 + \frac{1}{x^4})^{3/4}} dx$. 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: $I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$. Substituting $t$ back: $I = -(1 + \frac{1}{x^4})^{1/4} + c = -(\frac{x^4 + 1}{x^4})^{1/4} + c = -\frac{(x^4 + 1)^{1/4}}{x} + c$.

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

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