<s>[INST] $\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove 
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
   x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\bigg)$$ 
 [/INST] This is a weird one. Put $t = \frac{1}{4x}(1-u^2)$. The limits become $1$ and $\sqrt{1-4x}$. Then
$$ \frac{dt}{1-t} = \frac{2u \, du}{1-4x-u^2}, $$
while $1/\sqrt{1-4tx} = 1/u$.  But $1-4x-u^2 = (\sqrt{1-4x}+u)(\sqrt{1-4x}-u)$, so we can cancel part of this with the numerator of
$$\frac{1}{\sqrt{1-4x}}-1/u = \frac{u-\sqrt{1-4x}}{u\sqrt{1-4x}},$$
and we end up with the easily integrable
$$ \frac{2}{\sqrt{1-4x}}\int_{\sqrt{1-4x}}^1 \frac{du}{\sqrt{1-4x}+u}, $$
which gives
$$ \frac{2}{\sqrt{1-4x}} \left( \log{\left(1+\sqrt{1-4x}\right)} - \log{\left(2\sqrt{1-4x}\right)} \right) $$
as expected.
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