<s>[INST] Log integrals III The integral
\begin{align}
J_{m} = \int_{0}^{1} \frac{t^{m}}{1+t} \, \ln(1+t) \, dt 
\end{align}
has the general form 
\begin{align}
J_{m} = (-1)^{m} \left[ A_{m} - B_{m} \, \ln(2) + C_{m} \, \ln^{2}(2) \right]
\end{align}.
Is there a general form for the coefficients $A_{m}$, $B_{m}$, and $C_{m}$ ?
 [/INST] The generating function is, according to Maple, 
$$ \sum_{m=0}^\infty J_m x^m = \int_0^1 \dfrac{\ln(1+t)}{(1+t)(1-xt)}\; dt
= (1+x)^{-1} \left(\dfrac{1}{2} \ln(2)^2 + \ln \left(\dfrac{x}{1+x}\right) \ln(1-x) - \text{dilog} \left(\dfrac{x}{1+x}\right) + \text{dilog} \left(\dfrac{2x}{1+x}\right) \right)$$
but I don't know if you can get closed forms from that.
It appears (using the gfun package) that $(-1)^n A_n$ satisfies the recurrence
$$ \left( -{n}^{2}-5\,n-6 \right) a \left( n+1 \right) + \left( -{n}^{2}
-4\,n-3 \right) a \left( n+2 \right) + \left( {n}^{2}+9\,n+19 \right) 
a \left( n+3 \right) + \left( {n}^{2}+8\,n+16 \right) a \left( n+4
 \right) =0$$
while $(-1)^{n+1} B_n$ satisfies
$$  \left( -n-2 \right) b \left( n+1 \right) + \left( -n-2 \right) b
 \left( n+2 \right) + \left( n+4 \right) b \left( n+3 \right) +
 \left( n+4 \right) b \left( n+4 \right) =0$$
and $C_n = 1/2$.
</s>