The value of int frac{d x}{x^2left(x^4+1right | vedclass

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(C) Let $I = \int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$.Take $x^4$ common from the term inside the bracket:$I = \int \frac{d x}{x^2 \left[x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$.Take $x^4$ common from the term inside the bracket:$I = \int \frac{d x}{x^2 \left[x^4 \left(1 + \frac{1}{x^4}\right)\right]^{\frac{3}{4}}} = \int \frac{d x}{x^2 \cdot x^3 \left(1 + \frac{1}{x^4}\right)^{\frac{3}{4}}} = \int \frac{d x}{x^5 \left(1 + \frac{1}{x^4}\right)^{\frac{3}{4}}}$.Let $1 + \frac{1}{x^4} = t$. Then,differentiating both sides with respect to $x$:$-\frac{4}{x^5} dx = dt \Rightarrow \frac{dx}{x^5} = -\frac{dt}{4}$.Substituting these into the integral:$I = \int \left(1 + \frac{1}{x^4}\right)^{-\frac{3}{4}} \frac{dx}{x^5} = \int t^{-\frac{3}{4}} \left(-\frac{dt}{4}\right) = -\frac{1}{4} \int t^{-\frac{3}{4}} dt$.$I = -\frac{1}{4} \left[ \frac{t^{-\frac{3}{4} + 1}}{-\frac{3}{4} + 1} \right] + c = -\frac{1}{4} \left[ \frac{t^{\frac{1}{4}}}{\frac{1}{4}} \right] + c = -t^{\frac{1}{4}} + c$.Substituting $t = 1 + \frac{1}{x^4}$ back:$I = -\left(1 + \frac{1}{x^4}\right)^{\frac{1}{4}} + c = -\left(\frac{x^4 + 1}{x^4}\right)^{\frac{1}{4}} + c = -\frac{\left(x^4 + 1\right)^{\frac{1}{4}}}{x} + c$.

The value of $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

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