<s>[INST] Proving integration formula I want to prove the integration formula
$$
\int \frac {\sqrt {a+bu}}{u} \ du = 2 \sqrt {a+bu}+a \int { \frac {du}{u \sqrt {a+bu}}  }.
$$
I tried trigonometric substitution (as $u= \frac {a \tan^2 \theta}{b}$), but it didn't seem to work. I am stuck on how to prove the integration formula.
 [/INST] Let's try another substitution.
Let $x=\sqrt{a+bu}\Rightarrow u=\dfrac{x^2-a}{b} \qquad dx=b/2(a+bu)^{-1/2}du=\dfrac{bdu}{2x}\Rightarrow \dfrac{2xdx}{b}=du$ 
Now we have
${\Large{\int}} \dfrac{2x^2}{b}\cdot\dfrac{b}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2-(x^2-a)}{x^2-a}+\dfrac{x^2-a}{x^2-a}dx=2{\Large{\int}} \dfrac{a}{x^2-a}+1dx$
$=2x+2{\Large{\int}} \dfrac{a}{a+bu-a}b/2(a+bu)^{-1/2}du=2\sqrt{a+bu}+a{\Large{\int}} \dfrac{du}{u\sqrt{a+bu}}$
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