Find the value of the integral Find the value of the following integral conatining a term with natural logarithm$$\int_0^1 (1-y) \ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\, dy.$$
 A: Rewrite $\displaystyle\int_0^1 (1-y)\ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\ dy$ as
$$
\int_0^1 (1-y) \ln\left(2+\sqrt{1-y}\right)\ dy-\int_0^1 (1-y) \ln\left(2-\sqrt{1-y}\right)\ dy
$$
then let $x=2+\sqrt{1-y}$ and $z=2-\sqrt{1-y}$. The integral becomes
$$
\int_0^1 (1-y) \ln\left(2+\sqrt{1-y}\right)\ dy=2\int_2^3 (x-2)^3\ \ln x\ dx
$$
and
$$
\int_0^1 (1-y) \ln\left(2-\sqrt{1-y}\right)\ dy=-2\int_1^2 (z-2)^3\ \ln z\ dz,
$$
then
$$
\int_0^1 (1-y)\ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\ dy=2\int_1^3 (x-2)^3\ \ln x\ dx
$$
The last part can be solved using IBP by letting $u=\ln x$ and $dv=(x-2)^3$.
A: Hint
Separate the integral into two pieces (developing the logarithm of the ratio) and, say for the first one, change variable $$\sqrt{1-y}+2=e^z$$. You should arrive to something simple which only involves $z$ and some $e^{kz}$.
I am sure that you can take from here.
A: $$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy$$
Substitute $x=1-y$.
$$I=\int_{0}^{1}(1-y)\log{\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)}\,dy=\int_{0}^{1}x\,\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}\,dx.$$
This can be tackled by integration by parts. First we'll need the derivative of the logarithmic factor:
$$\frac{d}{dx}\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}=\left(\frac{2-\sqrt{x}}{2+\sqrt{x}}\right)\cdot\left(\frac{(2-\sqrt{x})\frac{1}{2\sqrt{x}}+(2+\sqrt{x})\frac{1}{2\sqrt{x}}}{(2-\sqrt{x})^2}\right)=\frac{2}{(4-x)\sqrt{x}}.$$
Then,
$$I=\frac12x^2\log{\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}\right)}\big |_{0}^{1}-\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx\\
=\frac12\log{3}-\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx.$$
To find the integral $\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx$, substitute $x=4u^2$: 
$$\int_{0}^{1}\frac{x^{3/2}}{4-x}\,dx=16\int_{0}^{\frac12}\frac{u^4}{1-u^2}\,du.$$
This last integral can readily be computed by expanding the integrand by partial fractions* and integrating term by term.

*Partial fraction expansion of $\frac{u^4}{1-u^2}$:
$$\frac{u^4}{1-u^2}=-u^2+\frac{u^2}{1-u^2}=-u^2-1+\frac{1}{1-u^2}=-u^2-1+\frac{1}{2(1+u)}+\frac{1}{2(1-u)}.$$
