int frac{x^4 cos left(tan ^{-1} x^5right)}{1+ | vedclass

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(A) Let $I = \int \frac{x^4 \cos \left(	an ^{-1} x^5ight)}{1+x^{10}} \,d x$.Substitute $u = 	an^{-1}(x^5)$.Then,$du = \frac{1}{1+(x^5)^2} \cdot \f

Vedclass · Accepted Answer

$\frac{\sin \left(	an ^{-1} x^5ight)}{5}+c$,where $c$ is the constant of integration

Vedclass · Answer

(A) Let $I = \int \frac{x^4 \cos \left(	an ^{-1} x^5ight)}{1+x^{10}} \,d x$.Substitute $u = 	an^{-1}(x^5)$.Then,$du = \frac{1}{1+(x^5)^2} \cdot \frac{d}{dx}(x^5) \,dx = \frac{5x^4}{1+x^{10}} \,dx$.This implies $\frac{x^4}{1+x^{10}} \,dx = \frac{du}{5}$.Substituting these into the integral,we get:$I = \int \cos(u) \cdot \frac{du}{5} = \frac{1}{5} \int \cos(u) \,du$.$I = \frac{1}{5} \sin(u) + c$.Substituting back $u = 	an^{-1}(x^5)$,we get:$I = \frac{\sin \left(	an ^{-1} x^5ight)}{5} + c$.

$\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} \,d x$ equals

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