A quick question on the integral of $\frac{\sqrt{x}}{\sqrt{x}-3}$

Question

Consider the function $$\int \frac{\sqrt{x}}{\sqrt{x}-3}$$ Using substitution we get $\begin{align*} u&= \sqrt{x}-3\\ x&=(u+3)^2 &dx= 2(u+3)du \end{align*}$

$(u+3)^2=$$u^2+6u+9$, plugging all this in we get $$\int\frac{u^2+6u+9}{u}du $$ This begin an even numbered question in my textbook, I used a online calculator which gave me $$\int\frac{2(u^2+6u+9)}{u}du$$ and I don't know where that $2$ came from. I think it is either from $dx$ but that would mean also having $(u+3)$ in there, or it could have come from $du=\frac{1}{2\sqrt{x}}$ but then where did that $\sqrt{x}$ go?

Answer 1

$\int \frac {\sqrt x} {\sqrt x-3}dx=\int \frac {u+3} {u} 2(u+3)du$ because $dx =2(u+3)du$. You just missed the $2$ in $dx$.

Answer 2

Try rewriting $\frac{\sqrt{x}}{\sqrt{x} - 3}$ as $1 + \frac{3}{\sqrt{x} - 3}$, then make the substitution $u = \sqrt{x}$ when integrating the second term. It should work out fine.

Answer 3

I agree with the on line calculator:

With the substitution $\color{blue}{u= \sqrt{x}-3}$, $\color{green}{\sqrt x=u+3}$, $x=(u+3)^2$, and $\color{red}{dx=2(u+3)du},$

$$\int \frac{\color{green}{\sqrt{x}}}{\color{blue}{\sqrt{x}-3}} \color{red}{dx}$$ becomes

$$\int\frac{\color{green}{u+3}}{\color{blue}u}\color{red}{\underline2(u+3)du}. $$

Answer 4

Notice that: $$\frac{\sqrt{x}}{\sqrt{x}-3}=1+\frac{3}{\sqrt{x}-3}$$ so our integral becomes: $$2\int\frac{u+3}{u}du=2(u+3\ln(u))+C$$