How can I integrate $\frac{1}{x^2-x-1}$? I need to find $$\int\frac{1}{x^2-x-1}dx$$ but I don't know what to do. I've thought about substitution or partial fractions but neither has worked.

As stated above, complete the square of the integrand to get 
\begin{equation*}
\int\frac{1}{(x-\frac{1}{2})^2-\frac{5}{4}}.
\end{equation*}
Now substitute $u=x-\frac{1}{2}$ & $du=dx:$
\begin{equation*}
\int\frac{1}{u^2-\frac{5}{4}}=-\frac{4}{5}\int \frac{1}{1-\frac{4u^2}{5}}.
\end{equation*}
Substitute again using $s=\frac{2u}{\sqrt{5}}$ & $ds=\frac{2}{\sqrt{5}}du:$
\begin{equation*}
-\frac{2}{\sqrt{5}}\int\frac{1}{1-s^2}ds=-\frac{2\tanh^{-1}(s)}{\sqrt{5}}+C.
\end{equation*}
Now substitute back for $s=\frac{2u}{\sqrt{5}}$ & $u=x-\frac{1}{2}.~_{\square}$
