<s>[INST] How do i find this : $\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$, where $a > b > 0$? Is there someone show me how do I find : $$\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$$, where $$a > b > 0$$ ?
I tried to make it as sum of fraction to be easier but sorry i didn't up
Thank you for any help .
 [/INST] Use the substitution $u^2 = x + b$.
Let 
$I = \int\frac{1}{(x+ a)\sqrt{x + b}}\,dx
$
Let
\begin{align*}
&u^2 = x + b\implies x + a = u^2 + a - b\\
&2u\,du = dx
\end{align*}
Then
\begin{align*}
I &= 2\int\frac{1}{u^2 + a - b}\,du\\
\\
&= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{u}{\sqrt{a - b}}\right) + C\\
\\
&= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{\sqrt{x + b}}{\sqrt{a - b}}\right) + C
\end{align*}
</s>