Integral of a log reciprocal I want to calculate following definite integral:
$$\int\limits_{-1}^0\left(\frac{1}{x\ln(1 + x)}-\frac{1}{x^2}-\frac{1}{2x}\right)dx$$
No idea from where to begin.
I know result is:
$$-\frac{1}{2}-\frac{\gamma_0}{2}+\frac{\ln(2\pi)}{2}$$
 A: The formula to be proved is (this corrects a sign error in the original statment)
$$ \int_{-1}^0 \frac{dx}{x^2}\Big( \, \frac{x}{\log{(1+x)}} -1 -\frac{x}{2} \Big) = 
\frac{1}{2}\big(1+\gamma - \log{(2 \pi)} \big).$$
The expression within the parentheses within the integral can be written in terms of the Bernoulli polynomials of the second kind (see the wiki)
$$ \frac{x}{\log{(1+x)}} -1 -\frac{x}{2}  = \sum_{n=2}^\infty a_n \, x^n \quad,\quad 
a_n = \int_{0}^1 \binom{u}{n} du.$$ 
Yes, that is a binomial within the integral. Also, $a_0=1, a_1=1/2.$ The radius of convergence is $|x|<1$ and since the integrand no longer has any problems within its domain, we can switch integral with sum and integrate right on up to -1.  
$$ \int_{-1}^0 \frac{dx}{x^2}\Big( \, \frac{x}{\log{(1+x)}} -1 -\frac{x}{2} \Big) = 
\sum_{n=0}^\infty a_{n+2} \int_{-1}^0 x^n\, \frac{dx}{x^2}  $$ 
$$ =\sum_{n=0}^\infty a_{n+2}\frac{(-1)^n}{n+1} = \int_0^1 du \sum_{n=0}^\infty \binom{u}{n+2}\frac{(-1)^n}{n+1} $$
$$ =  \int_0^1 du \, u\big(-1+\gamma + \psi(1+u) \big) $$
$$ = \frac{1}{2}\big(1+\gamma - \log{(2\pi)} \big)$$
In the penultimate line and last line I let Mathematica do the work for me, although I think the series and integral, which involves the polygamma function $\psi$, have been collected in tables of integral and series before.
