Question

$\int \frac{dx}{\sqrt{x}(x+9)}$ is equal to, $(x>0)$ (where $C$ is integration constant)

• A. $\frac{2}{9}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
• B. $2{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
• C. $\frac{2}{3}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$
• D. ${\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$

Answer 1

The correct option is C $\frac{2}{3}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$

$I=\int \frac{dx}{\sqrt{x}(x+9)}=\int \frac{dx}{\sqrt{x}({\left(\sqrt{x}\right)}^2+3^2)}$
Putting $\sqrt{x}=t$
$\Rightarrow \frac{dx}{\sqrt{x}}=2dt\Rightarrow I=\int \frac{2}{t^2+3^2}dt=\frac{2}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+C\{\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\}=\frac{2}{3}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+C$