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If $\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m$, then $(k, m)$ is
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Verified Answer
The correct answer is:
$(2, \sqrt{\mathrm{x}})$
$\begin{aligned}
&\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m \\
&\text { Put } \quad\left\{\begin{array}{l}
x=\tan ^{2} \theta \\
d x=2 \tan \theta \cdot \sec ^{2} \theta d \theta
\end{array}\right. \\
&\quad=\int \frac{\tan \theta}{\tan ^{2} \theta \cdot \sec ^{2} \theta} \cdot\left(2 \tan \theta \cdot \sec ^{2} \theta\right) d \theta \\
&=2 \int d \theta=2 \theta=2 \tan ^{-1} \sqrt{x}=k \tan ^{-1}(m)
\end{aligned}$
On comparing, we get $(\mathrm{k}, \mathrm{m})=(2, \sqrt{\mathrm{x}})$
&\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m \\
&\text { Put } \quad\left\{\begin{array}{l}
x=\tan ^{2} \theta \\
d x=2 \tan \theta \cdot \sec ^{2} \theta d \theta
\end{array}\right. \\
&\quad=\int \frac{\tan \theta}{\tan ^{2} \theta \cdot \sec ^{2} \theta} \cdot\left(2 \tan \theta \cdot \sec ^{2} \theta\right) d \theta \\
&=2 \int d \theta=2 \theta=2 \tan ^{-1} \sqrt{x}=k \tan ^{-1}(m)
\end{aligned}$
On comparing, we get $(\mathrm{k}, \mathrm{m})=(2, \sqrt{\mathrm{x}})$
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