TITLE: Evaluating the definite integral $\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $
QUESTION [5 upvotes]: So the question is to evaluate the following definite integral 

$$\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $$

My approach is very close to get answer, but I couldn't finish.
I assumed 
$$f(a) =\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $$ Now we've $$f'(a)= \int_0^u \frac{1}{ax} \cdot \frac{ x}{\sqrt{u-x}} \rm dx \implies f'(a) = \frac 1a \int_0^u \frac{1}{\sqrt{u-x}} \rm dx$$
$$f'(a) =-\frac{2 \sqrt{u-x}}{a} \Bigg |_0^u=\frac{2\sqrt u} a$$
$$\implies f(a) = \int \frac{2\sqrt u}{a}  \rm da = 2\sqrt u (\ln (a)+C)$$
Here $C = g(u)$ is possible. But I can't find out a way to reach to the final answer. 
Answer is $$\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx =2 \sqrt u ( \ln (4au)-2)$$
Can someone tell me how to proceed and/or can help evaluating this using any other method? 
Thanks.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{u}{\ln\pars{ax} \over \root{u - x}}\,\dd x & =
\int_{0}^{u}{\ln\pars{au\bracks{x/u}} \over \root{u}\root{1 - x/u}}\,u
\,{\dd x \over u} =
\root{u}\int_{0}^{1}\ln\pars{aux}\pars{1 - x}^{-1/2}\,\dd x
\\[5mm] & =
\root{u}\,\left.\partiald{}{z}\int_{0}^{1}\pars{aux}^{\,\large z}
\,\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ z\ =\ 0}
\\[5mm] & =
\root{u}\,\partiald{}{z}\bracks{\pars{au}^{\large\, z}\int_{0}^{1}x^{\,\large z}
\,\pars{1 - x}^{-1/2}\,\dd x}_{\ z\ =\ 0}
\\[5mm] & =
\root{u}\,\partiald{}{z}\bracks{\pars{au}^{\large\, z}\,
{\Gamma\pars{z + 1}\Gamma\pars{1/2} \over \Gamma\pars{z + 3/2}}}_{\ z\ =\ 0} =
\bbx{2\root{u}\bracks{\ln\pars{4au} - 2}}
\end{align}