If I=int frac{dx}{x^2(x^4+1)^{3/4}},then I is | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.We can rewrite the integral by taking $x^4$ common from the bracket:$I = \int \frac{dx}{x^2 \left[x^4(1 + \frac{1}{x^4})\right]^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.Let $t = 1 + \frac{1}{x^4}$. 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 back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:$I = -(1 + \frac{1}{x^4})^{1/4} + c = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c = -\frac{(x^4+1)^{1/4}}{x} + c$.

If $I=\int \frac{dx}{x^2(x^4+1)^{3/4}}$,then $I$ is

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