integration - Prove $int_0^1 frac{x-1}{(x+1)log{x}} text{d}x = log{frac{pi}{2}}$

Prove $$\int_0^1 \frac{x-1}{(x+1)\log{x}} \text{d}x = \log{\frac{\pi}{2}}$$

Tried contouring but couldn't get anywhere with a keyhole contour.

Geometric Series Expansion does not look very promising either.

Answer

Hint. One may set $$f(s):=\int_0^1 \frac{x^s-1}{(x+1)\log{x}}\: \text{d}x, \quad s>-1, \tag1$$ then one is allowed to differentiate under the integral sign, getting $$f'(s)=\int_{0}^{1}\frac{x^s}{x+1}\:dx=\frac12\psi\left(\frac{s}2+\frac12\right)-\frac12\psi\left(\frac{s}2+1\right), \quad s>-1, \tag2$$ where we have used a standard integral representation of the digamma function.

One may recall that $\psi:=\Gamma'/\Gamma$, then integrating $(2)$, observing that $f(0)=0$, one gets

$$f(s)=\int_0^1 \frac{x^s-1}{(x+1)\log{x}}\: \text{d}x=\log \left(\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{s}2+1\right)}{\Gamma\left(\frac{s}2+\frac12\right)}\right), \quad s>-1, \tag3$$

from which one deduces the value of the initial integral by putting $s:=1$, recalling that $$\Gamma\left(\frac12+1\right)=\frac12\Gamma\left(\frac12\right)=\frac{\sqrt{\pi}}2.$$

Edit. The result $(3)$ is more general than the given one.