How to evaluate $\int_{0}^{1} \frac{\ln x}{x+c} dx$ For $c=-1$ , it can be evaluated using the Taylor Series for $\ln x$ centered at $1$ to get $\zeta (2)$. For $c=1$ you enforce the substitution $x=1-u$ then use the Taylor series centered at $0$. 

Can we generalize for $c \in \mathbb{R}$?  

I've completed calc 1 and 2 but currently taking no math classes besides Ap statistics. I try to use this website to expand my math toolbox and learn tricks here and there. So can someone incorporate a technique that's not to far away for me to understand. 
Thanks.
Idea for $c<0$:
Taylor series for $\ln x$ about $x=-c$ is:
$$\ln (x)= \ln (-c)+ \sum_{n=1}^{\infty} \frac{(n-1)!(-1)^{n+1}}{(-c)^n} \frac{(x+c)^n}{n!}$$
I'm good from there... divide by $(x+c)$ and integrate.. and I'm left with a pretty weird sum, which I see as acceptable. Here's what I've got from this method:
$$\ln(-c) \int_{0}^{1} \frac{dx}{x+c}-\sum_{n=1}^{\infty} \frac{(1+c)^n-c^n}{n^2c^n}$$
Which can be simplified if $c \notin (-1,0)$
Now all that is left is $c \geq 0$
 A: I poked at this for a while using manual techniques and made (very) little progress.  So I asked a large computed algebra system (Mathematica 10.4.1) for an evaluation.  Here's what it reports.  For any $c \in \Bbb{C} \setminus (-1,0]$, 
\begin{align}
\int_0^1 \frac{\ln x}{x+c} \mathrm{d}x &= \frac{1}{6} \left( -\pi^2 - 3\left( \ln(-1-c) - \ln(-c) \right) \phantom{\left( \frac{c}{1+c} \right)}\right. \\ 
&\left. \left( \ln(-1-c) - \ln(-c) + 2 \ln\left(  \frac{1}{1+c}\right) \right) \right) +\mathrm{Li}_2\left( \frac{c}{1+c} \right)   \text{,}
\end{align} where $\mathrm{Li_2}$ is the dilogarithm.  (See also here, and Spence's function.)
You may look at those logs and think "Those simplify, obviously."  Yes, they do, but differently for different values of $c$.
\begin{align}
c &> 0 & \frac{1}{6} &\left(- \pi^2 -3 \ln^2(c)+3 \ln^2(c+1)  \right) + \mathrm{Li}_2 \left( \frac{c}{c+1}\right)  \\
c &\leq -1 & \frac{1}{6} &\left(-\pi ^2 -3 \ln\left(\frac{1}{c}+1\right) \left(\ln \left(\frac{1}{c}+1\right)+2 \ln\left(\frac{1}{c+1}\right)\right)\right) + \mathrm{Li}_2\left(\frac{c}{c+1}\right)  \\
c &\not \in \Bbb{R} & &\text{use the above long version}
\end{align}
How do we do the integral?  Beats me.  I see repeated logarithms, suggesting a particular pattern of substitutions, but not strongly enough to make the solution method jump out.
A: The Dilogarithm function can be defined as $$-\int_{0}^{1}\frac{\log\left(1-xt\right)}{x}dx=\textrm{Li}_{2}\left(t\right)
 $$ so in your case we have, taking $t=-\frac{1}{c},c\notin\left(-1,0\right]
 $ and integrating by parts, $$\textrm{Li}_{2}\left(-\frac{1}{c}\right)=-\int_{0}^{1}\frac{\log\left(1+x/c\right)}{x}dx$$ $$=-\left[\log\left(x\right)\log\left(1+x/c\right)\right]_{0}^{1}+\int_{0}^{1}\frac{\log\left(x\right)}{x+c}dx=\int_{0}^{1}\frac{\log\left(x\right)}{x+c}dx.$$
