int frac{x^2 tan^{-1}(x^3)}{1 + x^6} , dx is | vedclass

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Question

(B) Let $I = \int \frac{x^2 	an^{-1}(x^3)}{1 + x^6} \, dx$. Substitute $t = x^3$,then $dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$

Vedclass · Accepted Answer

$\frac{1}{6}(	an^{-1}(x^3))^2 + c$

Vedclass · Answer

(B) Let $I = \int \frac{x^2 	an^{-1}(x^3)}{1 + x^6} \, dx$. Substitute $t = x^3$,then $dt = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} \, dt$. The integral becomes $I = \frac{1}{3} \int \frac{	an^{-1}(t)}{1 + t^2} \, dt$. Now,substitute $z = 	an^{-1}(t)$,then $dz = \frac{1}{1 + t^2} \, dt$. Substituting these into the integral,we get $I = \frac{1}{3} \int z \, dz$. Integrating with respect to $z$,we get $I = \frac{1}{3} \cdot \frac{z^2}{2} + c = \frac{z^2}{6} + c$. Substituting back $z = 	an^{-1}(x^3)$,we get $I = \frac{1}{6}(	an^{-1}(x^3))^2 + c$.

$\int \frac{x^2 \tan^{-1}(x^3)}{1 + x^6} \, dx$ is equal to

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