int frac{x^3 tan^{-1} x^4}{1+x^8} dx = | vedclass

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(A) Let $I = \int \frac{x^3 	an^{-1} x^4}{1+x^8} dx$. Substitute $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting

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$\frac{(	an^{-1}(x^4))^2}{8} + c$

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(A) Let $I = \int \frac{x^3 	an^{-1} x^4}{1+x^8} dx$. Substitute $t = x^4$,then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting these into the integral,we get $I = \frac{1}{4} \int \frac{	an^{-1} t}{1+t^2} dt$. Now,let $u = 	an^{-1} t$,then $du = \frac{1}{1+t^2} dt$. Substituting $u$ into the integral,we get $I = \frac{1}{4} \int u du$. Integrating with respect to $u$,we get $I = \frac{1}{4} \cdot \frac{u^2}{2} + c = \frac{u^2}{8} + c$. Substituting back $u = 	an^{-1} t$ and $t = x^4$,we get $I = \frac{(	an^{-1}(x^4))^2}{8} + c$.

$\int \frac{x^3 \tan^{-1} x^4}{1+x^8} dx =$

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