Proving integration formula

Question

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.

Answer 1

First note that $\int \frac{\sqrt{a+bu}}{u} du=\int \frac{a+bu}{u\sqrt{a+bu}} du=b\int \frac{1}{\sqrt{a+bu}} du+a\int \frac{du}{u\sqrt{a+bu}} $

Now for the first term substitute $x=a+bu$ and it becomes $\int \frac{1}{\sqrt{x}} dx =2\sqrt{x}=2\sqrt{a+bu}$ And we're done.

Answer 2

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

${\Large{\int}} \dfrac{2x^2}{b}\cdot\dfrac{b}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2}{x^2-a}dx=2{\Large{\int}} \dfrac{x^2-(x^2-a)}{x^2-a}+\dfrac{x^2-a}{x^2-a}dx=2{\Large{\int}} \dfrac{a}{x^2-a}+1dx$

$=2x+2{\Large{\int}} \dfrac{a}{a+bu-a}b/2(a+bu)^{-1/2}du=2\sqrt{a+bu}+a{\Large{\int}} \dfrac{du}{u\sqrt{a+bu}}$