Solving integral of $\ln(A)/\sqrt{A}$ I am working on a physics problem that requires solving the following integral:
$$\int \frac{\ln(r^2+R^2+2rRu)\,du}{(r^2+R^2+2rRu)^{\frac{1}{2}}}$$
However, I don't know how to approach it, as it is unlike any I have seen before. I suspect that maybe the correct variable change will help me find it in an integral table, but I'm not sure.
 A: Once you get to the form of the integral in the title, just do integration by parts:$$\int fg' dA=fg-\int f'g dA$$
Use $f=\ln A$, with $f'=\frac 1A$, and $g'=A^{-1/2}$, with $g=2\sqrt A$
A: Consider the substitution $s=r^2+2rRu+R^2$ and thus $ds=2rRdu$. Then, the integral becomes
$$
\frac{1}{2rR}\int \frac{\log(s)}{\sqrt{s}} \,ds.
$$
Integrating by parts then leads to
$$
\frac{\sqrt{s}\log(s)}{rR}-\frac{1}{rR}\int \frac{1}{\sqrt{s}} \,ds.
$$
Since the integral of $1/\sqrt{s}$ is $2\sqrt{s}$, we get
$$
\frac{\sqrt{s}\log(s)}{rR}-\frac{2\sqrt{s}}{rR}.
$$
Reverting the change of variables, we finally get
$$
I=\frac{\sqrt{r^2+2rRu+R^2}(\log(r^2+2rRu+R^2)-2)}{rR}.
$$
A: I would use the substitution
$$t=\sqrt{r^2+R^2+2rRu}\iff t^2=r^2+R^2+2rRu,\enspace t\ge 0$$
We deduce $\;2t\,\mathrm dt =2rR\,\mathrm du$, so $\;\mathrm du=\frac t{rR}\,\mathrm d t$ and the integral becomes a standard integral:
\begin{align}
\int \frac{\ln(r^2+R^2+2rRu)}{(r^2+R^2+2rRu)^{1/2}}\,\mathrm du&=\int\frac{2\ln t}{t}\frac t{rR}\,\mathrm d t=\frac2{rR}\int \ln t\,\mathrm dt \\
&=\frac 2{rR}(t\ln t-t)\\&=\frac{(r^2+R^2+2rRu)^{1/2}}{rR}\Bigl(\ln(r^2+R^2+2rRu)-2\Bigr).
\end{align} 
