calculus - 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 $c \geq 0$

Answer

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, $$\textrm{Li}_{2}\left(-\frac{1}{c}\right)=-\int_{0}^{1}\frac{\log\left(1+x/c\right)}{x}dx$$ $$=-\left[\log\left(x\right)\log\left(1+x/c\right)\right]_{0}^{1}+\int_{0}^{1}\frac{\log\left(x\right)}{x+c}dx=\int_{0}^{1}\frac{\log\left(x\right)}{x+c}dx.$$