Evaluate $\int_0^1 \frac{\ln(1+bx)}{1+x} dx $ 
What is $ \displaystyle\int_0^1 \frac{\ln(1+bx)}{1+x} dx $?

I call it $f(b)$ and differentiate with respect to be $b,$ a bit of partial fractions and the $x$ integral can be done. Then I integrate with respect to $b$ and get a bit lost.
Can some of the terms be expressed in terms of dilogarithms? I get lost in the details! Could we avoid all this just go straight to dilogarithms (with a cunning substitution)?

Let $x = \frac{1}{b} \frac{1-u}{u}$. The integration range $0\leqslant x \leqslant 1$ translates to $\frac{1}{1+b} \leqslant u \leqslant 1$, for $b>0$
$$ 
  \frac{\log(1+ b x)}{1+x}\mathrm{d} x \to \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} 
   = \frac{\log(u)}{u} \left(1 + \frac{(1-b) u}{1-(1-b) u} \right)
$$
Now
$$ \begin{eqnarray}
  \int \frac{\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} &\stackrel{\color\red{\text{by parts}}}{=}& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \int \frac{\log(1-(1-b)u)}{u} \mathrm{d} u \\
  &=& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \operatorname{Li_2}\left((1-b) u\right) + C
\end{eqnarray}
$$
where a definition of dilogarithm was used $\operatorname{Li_2}^\prime(z) = -\frac{\log(1-z)}{z}$. Now assuming $b$ is such that the anti-derivatives is smooth:
$$
\begin{eqnarray}
  \int_0^1 \frac{\log(1+b x)}{1+x} \mathrm{d} x &=&  \int\limits_{\frac{1}{1+b}}^1 \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} \\ 
&=& \operatorname{Li_2}\left(1-b\right) - \operatorname{Li_2}\left(\frac{1-b}{1+b}\right) + \frac{1}{2} \log^2\left(1+b\right) +\log\left(1+b\right) \log\left(\frac{2b}{1+b}\right)
\end{eqnarray}
$$
