How do I integrate $\ln(x)/(x-1)$ How do I integrate $\ln(x)/(x-1)$?
\begin{align}
f(x)&=\int\frac{\ln(x)}{x-1} \,dx\\
&=\ln(x)\ln|x-1|-\int\frac{\ln|x-1|}{x}\, dx\\
&=\ln(x)\ln|x-1|-\left(\ln|x-1|\ln(x)-\int\frac{\ln(x)}{x-1}\, dx\right)\\
&=\ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
2f(x)&=0\\
f(x)&=0
\end{align}
I end up getting this but I know its incorrect. How do I do solve this
 A: As was mentioned in the comments $$ \int \frac{\ln(x)}{x-1}\,dx=-\operatorname{Li}_{2}(1-x)+C$$ where $\operatorname{Li}_{2}(x)$ is the polylogarithm function.  There isn't a way of writing this in terms of elementary functions.  Your error was in the line
\begin{align}
f(x) &= \ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
\implies 2f(x) &= 0.
\end{align}
You dropped a minus sign, so you should have
\begin{align}
f(x) &= \ln(x)\ln|x-1|-(\ln|x-1|\ln(x)-f(x))\\
\implies f(x) &= \ln(x)\ln|x-1|-\ln|x-1|\ln(x)+f(x)\\
\implies 0 &= 0
\end{align}
i.e. you haven't actually simplified the original expression.
A: To handle this integral, we need to use polylogarithms where $|z:
$$
\operatorname{Li_s}(z)=\sum_{k=1}^\infty{z^k\over k^s}
$$
which satisfies the property
$$
\operatorname{Li}_{s+1}(z)=\int_0^z{\operatorname{Li}_{s}(t)\over t}\mathrm dt
$$
If we set $s=1$, we can use the power series of logarithm to get
$$
\operatorname{Li}_1(z)=-\log(1-z)
$$
As a result, we get
$$
-\operatorname{Li}_2(1-z)=\int_0^{1-z}{\log(1-t)\over t}\mathrm dt=\int_1^z{\log t\over t-1}\mathrm dt
$$
