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

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(D) 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

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

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(D) 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:$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \left( \frac{t^{1/4}}{1/4} \right) + c = -t^{1/4} + c$.Substituting back $t = 1 + \frac{1}{x^4}$:$I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + c$.

The value of $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$ is

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