Integrating Logs Looking at: $$\int\frac{\sqrt{x}}{\sqrt{x}-3}$$
I use these steps to find the integral of the equation
$$\int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}$$
$$\int\frac{\sqrt{x}-3}{\sqrt{x}-3}+3\int\frac{1}{\sqrt{x}-3}$$
Now substituting $u=\sqrt{x}-3$ and $du=\frac{1}{2\sqrt{x}}$
$$\int1+3(2)\int\frac{u+3}{u}$$
$$\int1\,dx+3(2)(3)\int\frac{1}{u}du+3(2)\int1\,du$$
$$x+3(2(3)\ln|u|+2u)$$
$$x+18\ln|u|+6u)$$
Ending up with
$$x+18\ln|\sqrt{x}-3|+6(\sqrt{x}-3)+C$$
My issue is with the substitution. I got $2\,du=\frac{1}{\sqrt{x}}\,dx$ or $2\sqrt{x}\,du=1\,dx$
So, if I look at $\int\frac{u+3}{u}$ $2\sqrt{x}\int\frac{\sqrt{x}}{\sqrt{x}-3}$ gives me what I want $2\int\frac{1}{\sqrt{x}-3}$. Am I thinking about this part correctly?

\begin{align}
u & = \sqrt x - 3 \\[10pt]
u - 3 & = \sqrt x\\[10pt]
(u-3)^2 & = x \\[10pt]
2(u-3)\,du & = dx \\[10pt]
\int \frac{dx}{\sqrt x - 3} & = \int \frac{2(u-3)\,du} u = 2 \int \left( \frac u u - \frac 3 u \right) \, du = 2\int\left( 1 - \frac 3 u \right)\, du = \text{etc.}
\end{align}
