Can the following integral $I(c): = \int_{|c-1|}^{c+1} \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx$ be computed? Let $c > 0$ be a real number and consider the following integral: 
\begin{equation} I(c): = \int_{|c-1|}^{c+1}  \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx \end{equation}
I make the following claim: 
\begin{align} I(c) = \begin{cases} \ln(c) & c > 1 \\ 0 & 0 < c \leq 1 \end{cases} \end{align}
Indeed, when $c > 1$, a calculation on wolfram alpha will always yield a decimal approximate for $\ln(c)$, and when $0 < c \leq 1$, one will obtain $0$ as the exact result the very same way. In fact, when $c \leq 1$, the corresponding integrand suddenly seems to become symmetric about $x = 1$. 
Let $c \leq 1$. Then, for every  $k \in [0,c)$, the integrand $F_c(x) := \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}}$ satisfies \begin{equation} F_c(1 + k) = - F_c(1 - k). \end{equation}
These are all quite peculiar equalities, as they are not apparent at all to me, and I am very much interested in how to prove them. Any help is appreciated. 
 A: 
Any help is appreciated.

Since $~2x=\big(x^2\big)',\quad\ln x=\dfrac{\ln\big(x^2\big)}2,\quad$ and $x^4=\big(x^2\big)^2,\quad$ substituting $x^2\mapsto x$ seems like the 
natural course of action. We then have $I(c)=\dfrac1{2\pi}\displaystyle\int_{(1-c)^2}^{(1+c)^2}\frac{\ln x~\cdot~dx}{\sqrt{-x^2+2(1+c^2)~x-(1-c^2)^2}}.$
Completing the square in the denominator, we have $P(x)=(2c)^2-\Big[x-(1+c^2)\Big]^2.$ Factoring 
$2c$ outside the radical, and substituting $\dfrac{x-(1+c^2)}{2c}\mapsto x,~$ we eventually obtain $I(c)=\dfrac{J(c)}{2\pi},$
where $J(c)=\displaystyle\int_{-1}^1\frac{\ln(c^2\pm2cx+1)}{\sqrt{1-x^2}}~dx.~$ Factoring again, and using the basic properties of the 
logarithm, we have $J(c)=\displaystyle\int_{-1}^1\frac{\ln(2c)}{\sqrt{1-x^2}}~dx~+~\int_{-1}^1\frac{\ln(x+a)}{\sqrt{1-x^2}}~dx,~$ where $a=\dfrac{1+c^2}{2c}\ge1$
for all $c\ge0.~$ The former is trivial to evaluate, yielding $J_1(c)=\pi\ln(2c).~$ As for the latter, by 
differentiating $K(b)=\displaystyle\int_{-1}^1\frac{\ln(a+bx)}{\sqrt{1-x^2}}~dx~$ under the integral sign with regard to b, we are left 
with $K'(b)=\displaystyle\int_{-1}^1\frac{x~\cdot~dx}{(a+bx)\sqrt{1-x^2}}=\frac\pi b\bigg(1-\frac a{\sqrt{a^2-b^2}}\bigg),~$ which, after being integrated 
back with regards to b, yields $K(1)-K(0)=\displaystyle\int_0^1K'(b)~db=\pi~\Big\{\mid\ln c\mid-\ln(2a)\Big\}.~$ Since 
$K(0)=\pi\ln a,~$ it follows that $J_2(c)=K(1)=\pi~\Big\{\mid\ln c\mid-\ln2\Big\},~$ and, by extension, that 
$I(c)=0$ for $c\in(0,1],~$ and $I(c)=\ln c,~$ for $c\ge1.$
