int frac{x^{49} tan ^{-1}left(x^{50}right)}{1 | vedclass

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(C) Let $I = \int \frac{x^{49} 	an ^{-1}\left(x^{50}ight)}{1+x^{100}} d x$.Substitute $t = x^{50}$, then $dt = 50x^{49} dx$, which implies $x^{49}

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$\frac{1}{100}$

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(C) Let $I = \int \frac{x^{49} 	an ^{-1}\left(x^{50}ight)}{1+x^{100}} d x$.Substitute $t = x^{50}$, then $dt = 50x^{49} dx$, which implies $x^{49} dx = \frac{1}{50} dt$.Substituting these into the integral, we get:$I = \frac{1}{50} \int \frac{	an ^{-1} t}{1+t^2} dt$.Now, let $u = 	an ^{-1} t$, then $du = \frac{1}{1+t^2} dt$.The integral becomes:$I = \frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + c = \frac{u^2}{100} + c$.Substituting back $u = 	an ^{-1} (x^{50})$, we have:$I = \frac{(	an ^{-1} (x^{50}))^2}{100} + c$.Comparing this with the given expression $k(	an ^{-1} (x^{50}))^2 + c$, we find $k = \frac{1}{100}$.

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k$ is equal to

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