int frac{3x^2}{x^6 + 1} dx = | vedclass

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Question

(B) Let $I = \int \frac{3x^2}{x^6 + 1} dx$. Substitute $t = x^3$. Then,differentiating both sides with respect to $x$,we get $dt = 3x^2 dx$. Substitut

Vedclass · Accepted Answer

$	an^{-1}(x^3) + c$

Vedclass · Answer

(B) Let $I = \int \frac{3x^2}{x^6 + 1} dx$. Substitute $t = x^3$. Then,differentiating both sides with respect to $x$,we get $dt = 3x^2 dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t^2 + 1} dt$. We know that $\int \frac{1}{t^2 + 1} dt = 	an^{-1}(t) + c$. Substituting $t = x^3$ back,we get $I = 	an^{-1}(x^3) + c$.

$\int \frac{3x^2}{x^6 + 1} dx = $

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