<s>[INST] How to find $\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$ Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.
From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boyadzhiev/boyadzhiev6.pdf
 , I found that $$\sum_{n=0}^\infty \binom{2n}{n}H_n x^n=\frac{1}{\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)=f(x)$$
So I get $A=\displaystyle\int_0^{1/4}\frac{f(x)}{x}dx$.
How to find the exact value of
$$\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$$.
Thank in advances.
 [/INST] Hint :
by $$u=\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}$$
$$\Rightarrow \sqrt{1-4x}=\frac{1}{2u-1}$$
$$\Rightarrow x=\frac{1}{4}(1-\frac{1}{(2u-1)^2})$$
$$dx=\frac{1}{(2u-1)^3}du$$
you get $$I=\int_1^{\infty}\frac{\ln u}{u(u-1)}du=-\sum_{n=0}^{\infty}\int_1^{\infty}u^{n-1}\ln u du=-\sum_{n=0}^{\infty}-\frac{1}{n^2}=\frac{\pi^2}{6}$$
for the last integral use integration by part
notice that $$\frac{1}{u-1}=-\sum_{n=0}^{\infty}u^n$$
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