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$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{100}$
Let $I=\int x^{49} \frac{\tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^2} d x$
Let $x^{50}=t \Rightarrow 50 x^{49} d x=d t$
$\therefore \quad I=\frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} d t$
Let $\tan ^{-1} t=u \Rightarrow \frac{1}{1+t^2} d t=d u$
$\begin{aligned}
\therefore \quad I & =\frac{1}{50} \int u d u=\frac{u^2}{100}+c \\
& =\frac{\left(\tan ^{-1} x^{50}\right)^2}{100}+c
\end{aligned}$
But $I=k\left(\tan ^{-1} x^{50}\right)^2+c$(given)
$\begin{array}{rlrl}\Rightarrow & k\left(\tan ^{-1} x^{50}\right)^2+c & =\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+c \\ \Rightarrow & & k & =\frac{1}{100}\end{array}$
Let $x^{50}=t \Rightarrow 50 x^{49} d x=d t$
$\therefore \quad I=\frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} d t$
Let $\tan ^{-1} t=u \Rightarrow \frac{1}{1+t^2} d t=d u$
$\begin{aligned}
\therefore \quad I & =\frac{1}{50} \int u d u=\frac{u^2}{100}+c \\
& =\frac{\left(\tan ^{-1} x^{50}\right)^2}{100}+c
\end{aligned}$
But $I=k\left(\tan ^{-1} x^{50}\right)^2+c$(given)
$\begin{array}{rlrl}\Rightarrow & k\left(\tan ^{-1} x^{50}\right)^2+c & =\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+c \\ \Rightarrow & & k & =\frac{1}{100}\end{array}$
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