If $x > 1$, prove that $f(x) = \dfrac{1}{\sqrt{x^2+1}}\log(x+\sqrt{x^2-1})$ $$f(x) = 2\int_{0}^{1}\dfrac{du}{u^2(1-x)+1+x}$$
I have used partial fractions but solves nothing.
 A: Since $x>1$ the denominator better reads as $(1+x)-u^2(x-1)$. Now, let's change variables $u = \sqrt{\frac{x+1}{x-1}} w$, giving:
$$
   f(x) = \frac{2}{1+x} \sqrt{\frac{x+1}{x-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \frac{\mathrm{d}w}{1-w^2} = \frac{1}{\sqrt{x^2-1}} \int_0^\sqrt{\frac{x-1}{x+1}} \left(\frac{1}{1+w} + \frac{1}{1-w}\right) \mathrm{d}w
$$
Giving
$$  
f(x) =  \frac{1}{\sqrt{x^2-1}} \log \frac{1+\sqrt{\frac{x-1}{x+1}} }{1-\sqrt{\frac{x-1}{x+1}}} = \frac{1}{\sqrt{x^2-1}} \log \left( \frac{\sqrt{x-1} + \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x-1}} \cdot \color\green{\frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} + \sqrt{x-1}}} \right) = \frac{1}{\sqrt{x^2-1}} \log \left( \frac{ (x+1)+(x-1)+2\sqrt{x^2-1} }{(x+1)-(x-1)} \right) = \frac{1}{\sqrt{x^2-1}} \log \left( x+\sqrt{x^2-1}  \right)
$$
A: I don't see a logarithm but an arctangent. Check this:
$$(1-x)u^2+1+x=(1+x)\left(\frac{1-x}{1+x}u^2+1\right)=(1+x)\left(1+\left(\sqrt\frac{1-x}{1+x}\;u\right)^2\right)\Longrightarrow$$
$$2\int\limits_0^1\frac{du}{(1-x)u^2+1+x}=\frac{2}{\sqrt{(1-x)(1+x)}}\;\int\limits_0^1\frac{\sqrt{\frac{1-x}{1+x}}\;du}{\left(1+\left(\sqrt\frac{1-x}{1+x}\;u\right)^2\right)}=$$
$$=\frac{2}{\sqrt{(1-x)(1+x)}}\arctan\left(\sqrt\frac{1-x}{1+x}\;u\right)_0^1=\frac{2}{\sqrt{(1-x)(1+x)}}\arctan\left(\sqrt\frac{1-x}{1+x}\right)$$
