int frac{x^2}{1+x^6} d x is equal to | vedclass

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(B) Let $I = \int \frac{x^2}{1 + (x^3)^2} dx$.Substitute $z = x^3$, then $dz = 3x^2 dx$, which implies $x^2 dx = \frac{1}{3} dz$.Substituting these in

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$\frac{1}{3} 	an ^{-1}\left(x^3ight)+C$

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(B) Let $I = \int \frac{x^2}{1 + (x^3)^2} dx$.Substitute $z = x^3$, then $dz = 3x^2 dx$, which implies $x^2 dx = \frac{1}{3} dz$.Substituting these into the integral, we get $I = \frac{1}{3} \int \frac{1}{1 + z^2} dz$.Using the standard integral formula $\int \frac{1}{1 + z^2} dz = 	an^{-1}(z) + C$, we obtain $I = \frac{1}{3} 	an^{-1}(z) + C$.Finally, substituting $z = x^3$ back, we get $I = \frac{1}{3} 	an^{-1}(x^3) + C$.

$\int \frac{x^2}{1+x^6} d x$ is equal to

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