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If $\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k=$
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Verified Answer
The correct answer is:
$\frac{1}{100}$
$\int \frac{x^{49} \tan ^{-1} x^{50}}{1+x^{100}} d x$
Let $\tan ^{-1} x^{50}=\mathrm{t}$
$$
\begin{aligned}
& \frac{50 x^{49}}{1+\left(x^{50}\right)^2} d x=d t \\
& \frac{1}{50} \int t d t=\frac{1}{50} \frac{t^2}{2}+\mathrm{C}=\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+\mathrm{C} \\
& \therefore k=\frac{1}{100}
\end{aligned}
$$
Let $\tan ^{-1} x^{50}=\mathrm{t}$
$$
\begin{aligned}
& \frac{50 x^{49}}{1+\left(x^{50}\right)^2} d x=d t \\
& \frac{1}{50} \int t d t=\frac{1}{50} \frac{t^2}{2}+\mathrm{C}=\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+\mathrm{C} \\
& \therefore k=\frac{1}{100}
\end{aligned}
$$
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