Q62 / 150 · MHT CET 2021 (24 Sep Shift 1) · Mathematics
If $\int \frac{\sqrt{x}}{x(x+1)} d x=k \tan ^{-1} m+c$, (where $c$ is constant of integration), then
- A$\mathrm{k}=1, \mathrm{~m}=\sqrt{\mathrm{x}}$
- Correct answerB$\mathrm{k}=2, \mathrm{~m}=\sqrt{\mathrm{x}}$
- C$\mathrm{k}=1, \mathrm{~m}=\mathrm{x}$
- D$\mathrm{k}=2, \mathrm{~m}=\mathrm{x}$
Solution
$$
I=\int \frac{\sqrt{x}}{x(x+1)} d x
$$
Put $x \tan ^2 \theta \Rightarrow d x=2 \tan \theta \sec ^2 \theta d \theta$
$$
\begin{aligned}
& \therefore I=\int \frac{\tan \theta\left(2 \tan \theta \sec ^2 \theta\right)}{\tan ^2 \theta(1+\tan \theta)} d \theta \\
& =2 \int \frac{\sec ^2 \theta}{\sec ^2 \theta} d \theta=2 \int d \theta=2 \theta \\
& =2 \tan ^{-1} \sqrt{x}+c
\end{aligned}
$$
Comparing with given data, $\mathrm{k}=2, \mathrm{~m}=\sqrt{\mathrm{x}}$
Hence, the correct answer is option: 'B'.