int frac{4 x^2 cot ^{-1}left(x^3right)}{1+x^6 | vedclass

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(B) Let $I = \int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x$. Substitute $t = \cot ^{-1}\left(x^3\right)$. Then,the derivative is $\frac{dt

Vedclass · Accepted Answer

$\frac{-2}{3}\left(\cot ^{-1} x^3\right)^2+C$

Vedclass · Answer

(B) Let $I = \int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x$. Substitute $t = \cot ^{-1}\left(x^3\right)$. Then,the derivative is $\frac{dt}{dx} = -\frac{1}{1+(x^3)^2} \cdot 3x^2 = -\frac{3x^2}{1+x^6}$. This implies $\frac{x^2}{1+x^6} dx = -\frac{1}{3} dt$. Substituting these into the integral: $I = 4 \int t \left(-\frac{1}{3} dt\right) = -\frac{4}{3} \int t \,dt$. Integrating with respect to $t$: $I = -\frac{4}{3} \cdot \frac{t^2}{2} + C = -\frac{2}{3} t^2 + C$. Substituting back $t = \cot ^{-1}\left(x^3\right)$: $I = -\frac{2}{3} \left(\cot ^{-1} x^3\right)^2 + C$.

$\int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x=$ (where $C$ is a constant of integration.)

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