If int x^{49} left[ operatorname{Tan}^{-1} x^ | vedclass

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Question

(A) Let $I = \int x^{49} \left[ \operatorname{Tan}^{-1} x^{50} + \frac{x^{50}}{1 + x^{100}} \right] dx$. Substitute $u = x^{50}$,then $du = 50x^{49} d

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$k+n$

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(A) Let $I = \int x^{49} \left[ \operatorname{Tan}^{-1} x^{50} + \frac{x^{50}}{1 + x^{100}} \right] dx$. Substitute $u = x^{50}$,then $du = 50x^{49} dx$,which implies $x^{49} dx = \frac{du}{50}$. The integral becomes $I = \frac{1}{50} \int \left( \operatorname{Tan}^{-1} u + \frac{u}{1 + u^2} \right) du$. Using integration by parts for $\int \operatorname{Tan}^{-1} u \, du = u \operatorname{Tan}^{-1} u - \int \frac{u}{1 + u^2} du$. Substituting this back,$I = \frac{1}{50} \left( u \operatorname{Tan}^{-1} u - \int \frac{u}{1 + u^2} du + \int \frac{u}{1 + u^2} du \right) = \frac{u \operatorname{Tan}^{-1} u}{50} + c$. Substituting $u = x^{50}$ back,$I = \frac{x^{50}}{50} \operatorname{Tan}^{-1} x^{50} + c$. Comparing this with $\frac{x^n}{k} f(x) + c$,we have $n = 50$,$k = 50$,and $f(x) = \operatorname{Tan}^{-1} x^{50}$. Then $f\left(\sqrt[k]{x^n}\right) = f\left(\sqrt[50]{x^{50}}\right) = f(x) = \operatorname{Tan}^{-1} x^{50}$. Thus,$f(x) - f\left(\sqrt[k]{x^n}\right) = \operatorname{Tan}^{-1} x^{50} - \operatorname{Tan}^{-1} x^{50} = 0$. Wait,checking the options provided,there seems to be a discrepancy in the question structure. Re-evaluating the expression $\frac{x^n}{k} f(x)$,if $f(x) = \operatorname{Tan}^{-1} x^{50}$,then $f(x) - f(x) = 0$. Given the options,if the question implies $f(x) = \operatorname{Tan}^{-1} x$,then $f(x) - f(x^{50/50}) = 0$. Since $0$ is not an option,let's re-examine the integral: $\int x^{49} \operatorname{Tan}^{-1} x^{50} dx + \int \frac{x^{99}}{1+x^{100}} dx = \frac{1}{50} x^{50} \operatorname{Tan}^{-1} x^{50} - \frac{1}{50} \int \frac{50x^{49} x^{50}}{1+x^{100}} dx + \int \frac{x^{99}}{1+x^{100}} dx = \frac{x^{50}}{50} \operatorname{Tan}^{-1} x^{50}$. The result is $0$.

If $\int x^{49} \left[ \operatorname{Tan}^{-1} x^{50} + \frac{x^{50}}{1 + x^{100}} \right] dx = \frac{x^n}{k} f(x) + c$,then $f(x) - f\left(\sqrt[k]{x^n}\right) =$

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