<s>[INST] Find the value of the integral Find the value of the following integral conatining a term with natural logarithm$$\int_0^1 (1-y) \ln\left(\frac{2+\sqrt{1-y}}{2-\sqrt{1-y}}\right)\, dy.$$
 [/INST] 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 $dv=(x-2)^3$.
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