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Question: Answered & Verified by Expert
$\int \frac{d x}{\sqrt{x}(x+9)}$ is equal to
MathematicsIndefinite IntegrationAP EAMCETAP EAMCET 2001
Options:
  • A $\frac{2}{3} \tan ^{-1} \sqrt{x}+C$
  • B $\frac{2}{3} \tan ^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
  • C $\tan ^{-1}(\sqrt{x})+C$
  • D $\tan ^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
Solution:
2678 Upvotes Verified Answer
The correct answer is: $\frac{2}{3} \tan ^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
Let
$I=\int \frac{d x}{\sqrt{x}(x+9)}$
Put
$\begin{aligned} x & =t^2 \Rightarrow d x=2 t d t \\ & =\int \frac{2 t d t}{t\left(t^2+9\right)}=2 \cdot \frac{1}{3} \tan ^{-1} \frac{t}{3}+C \\ & =\frac{2}{3} \tan ^{-1} \frac{\sqrt{x}}{3}+C\end{aligned}$

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