$\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove 
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
   x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\bigg)$$ 

Hint 1:
$$\sqrt{1-4xt}-\sqrt{1-4x}={4x(1-t)\over\sqrt{1-4xt}+\sqrt{1-4x}}$$
which suggests the change of variable $u=4xt$.
Hint 2:  Letting $1-u=w^2$ (and abbreviating $\sqrt{1-4x}$ to $a$), we have
$$\int{du\over a\sqrt{1-u}(\sqrt{1-u}+a)}={1\over a}\int{-2wdw\over w(w+a)}$$
Now use partial fractions.
