If int frac{x^{49} tan ^{-1}(x^{50})}{1+x^{10 | vedclass

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(D) Let $I = \int \frac{x^{49} 	an ^{-1}(x^{50})}{1+x^{100}} d x$. Substitute $t = 	an ^{-1}(x^{50})$. Then, differentiating both sides with respect

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$\frac{1}{100}$

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(D) Let $I = \int \frac{x^{49} 	an ^{-1}(x^{50})}{1+x^{100}} d x$. Substitute $t = 	an ^{-1}(x^{50})$. Then, differentiating both sides with respect to $x$, we get: $\frac{dt}{dx} = \frac{1}{1+(x^{50})^2} \cdot 50x^{49} = \frac{50x^{49}}{1+x^{100}}$. Thus, $\frac{x^{49}}{1+x^{100}} dx = \frac{1}{50} dt$. Substituting these into the integral: $I = \int t \cdot \frac{1}{50} dt = \frac{1}{50} \int t dt = \frac{1}{50} \cdot \frac{t^2}{2} + C = \frac{1}{100} t^2 + C$. Replacing $t$ back, we get $I = \frac{1}{100} (	an ^{-1}(x^{50}))^2 + C$. Comparing this with the given expression $k(	an ^{-1}(x^{50}))^2 + c$, we find $k = \frac{1}{100}$.

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

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