Question

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

• A. $\frac{2}{5}{\tan }^{-1}\left(\sqrt{x}\right)+c$
• B. $\frac{2}{3}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+c$
• C. ${\tan }^{-1}\left(\sqrt{x}\right)+c$
• D. ${\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)+c$

Answer 1

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

$\int \frac{dx}{\sqrt{x}(x+9)}\Rightarrow \sqrt{x}=U$

$\frac{1}{2\sqrt{x}}dx=dv\Rightarrow 2vdv=dx$

$\int \frac{dx}{U(U^2+9)}-\frac{2}{3}{\tan }^{-1}\left(\frac{U}{3}\right)=\frac{2}{3}{\tan }^{-1}\left(\frac{\sqrt{x}}{3}\right)$