Can the definite integral $ I(n) := \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2}dx$ be computed? For $n \in \mathbb N$, $n \geq 1$, consider the following integral expression:
\begin{equation} I(n) :=  \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx \end{equation}
My attempts to use any of the obvious (elementary) few methods that I am accustomed with (i.e partial integration, substitution, series expansion of the integrand) in order to find a general antiderivative were all futile. There might still be a way to compute an antiderivative using complex analysis, but this is currently beyond my capabilites. However, although Wolfram Alpha does also not seem to be able to either to compute a general antiderivate or compute the above expression for general $n \in \mathbb N$, it will yield (after possibly some refreshing) an explicit value for any $n$ I've tested so far (all $n$ between $1$ and $15$). Namely, one gets the following result for $1 \leq n \leq 15:$
\begin{equation} I(n)= (n-\frac{1}{2}) \ln(2n-1) - (n-1) \ln(2n). \end{equation}
This suggest that there is indeed a way to compute the above expression explicitly. Any help is highly appreciated.
 A: Hint. By making the change of variable, 
$$
x=2\sqrt{2n-1}\:u ,\quad dx=2\sqrt{2n-1}\:du,\quad u=\frac{x}{2\sqrt{2n-1}},
$$ one has
$$
\frac{\pi}{2n}\cdot I(n)=\ln(2\sqrt{2n-1})\int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du+\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du \tag1
$$ with
$$
a:=\frac{n}{\sqrt{2n-1}}>1, \quad (n>1).\tag2
$$ The first integral on the right hand side of $(1)$ may be evaluated by two changes of variable, $u=\sin \theta$ and $t=\tan\theta$, obtaining, for $a>1$,

$$
\int_0^1\frac{\sqrt{1-u^2}}{a^2-u^2}\:du=\frac12\int_0^\infty\frac{1}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt=\frac{\pi  \left(a-\sqrt{a^2-1}\right)}{2 a}.\tag3
$$ 

Similarly, the second integral on the right hand side of $(1)$ is such that
$$
\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du= \frac{1}{2}\int_0^\infty\frac{2\ln t-\ln(1+t^2)}{\left((a^2-1)t^2+a^2\right)\cdot\left(t^2+1\right)}\:dt,\quad (a>1),\tag4
$$ transforming the integrand by a partial fraction decomposition, using the standard results
$$
\begin{align}
\int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt&=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha},\quad \alpha>0, \qquad (\star)
\\\int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt&=\pi\cdot\frac{\ln(\alpha+1)}{\alpha},\quad \alpha>0,\qquad (\star \star)
\end{align}
$$ one gets

$$
\int_0^1\frac{\ln u \cdot\sqrt{1-u^2}}{a^2-u^2}\:du=\frac{\pi\sqrt{a^2-1}}{2 a} \cdot\ln\left(\frac{\sqrt{a^2-1}+a}{a}\right)-\frac{\pi}2 \ln 2,\quad (a>1).\tag5
$$ 

Combining $(3)$ and $(5)$ with $(2)$ gives

$$
I(n) :=  \frac{2n}{\pi}\int_0^{2\sqrt{2n-1}} \frac{\ln(x)\sqrt{-x^2 + 8n-4}}{4n^2-x^2} dx= \left(n-\frac{1}{2}\right) \ln(2n-1) - (n-1) \ln(2n)
$$ 

for all real numbers $n$ such that $n>1$.

Edit. To prove $(\star)$, one may perform the change of variable $t=\alpha u$,  $\alpha>0$, getting
$$
\int_0^\infty\frac{\ln t}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln \alpha+ \ln u}{u^2+1}\:du=\frac{\pi}{2}\cdot\frac{\ln\alpha}{\alpha}
$$ since
$$
\int_0^\infty\frac{1}{u^2+1}\:du=\left[\frac{}{}\arctan u \frac{}{}\right]_0^\infty=\frac \pi2, \quad \int_0^\infty\frac{\ln u}{u^2+1}\:du=0
$$ (the latter is seen by making $u \to 1/u$).
To prove $(\star\star)$, one may perform the change of variable $t=\alpha u$,  $\alpha>0$, getting
$$
\int_0^\infty\frac{\ln (1+t^2)}{t^2+\alpha^2}\:dt=\frac1\alpha \cdot\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du
$$ differentiating the latter integral with respect to $\alpha$ gives a classic evaluation
$$
\frac{d}{d\alpha}\int_0^\infty\frac{\ln (1+\alpha^2u^2)}{u^2+1}\:du=\int_0^\infty\frac{2\alpha u^2}{(u^2+1)(1+\alpha^2u^2)}\:du=\frac{\pi}{\alpha+1}
$$ then integrating one gets $(\star\star)$.
