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

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(A) Let $I = \int \frac{dx}{x^2(x^4 + 1)^{3/4}}$.Taking $x^4$ common from the term inside the bracket in the denominator:$I = \int \frac{dx}{x^2 \left

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x^2(x^4 + 1)^{3/4}}$.Taking $x^4$ common from the term inside the bracket in the denominator:$I = \int \frac{dx}{x^2 \left[ x^4(1 + \frac{1}{x^4}) \right]^{3/4}}$$I = \int \frac{dx}{x^2 \cdot (x^4)^{3/4} \cdot (1 + \frac{1}{x^4})^{3/4}}$$I = \int \frac{dx}{x^2 \cdot x^3 \cdot (1 + \frac{1}{x^4})^{3/4}}$$I = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$Let $t = 1 + \frac{1}{x^4}$. Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dt}{-4} = \frac{dx}{x^5}$.Substituting these into the integral:$I = \int \frac{1}{t^{3/4}} \cdot (-\frac{1}{4} dt)$$I = -\frac{1}{4} \int t^{-3/4} dt$$I = -\frac{1}{4} \left[ \frac{t^{1/4}}{1/4} \right] + c$$I = -t^{1/4} + c$Substituting $t = 1 + \frac{1}{x^4}$ back:$I = -\left( 1 + \frac{1}{x^4} \right)^{1/4} + c$$I = -\left( \frac{x^4 + 1}{x^4} \right)^{1/4} + c$.

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

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