How can I integrate $\frac{1}{x^2-x-1}$?

Question

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.

Answer 1

$$\dfrac1{x^2-x-1} = \dfrac1{(x-1/2)^2-(\sqrt5/2)^2} = \dfrac1{(x-a)(x-b)} = \dfrac1{a-b}\left(\dfrac1{x-a} - \dfrac1{x-b} \right)$$ where $a=\dfrac{1+\sqrt5}2$ and $b = \dfrac{1-\sqrt5}2$.

Answer 2

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 $s=\frac{2u}{\sqrt{5}}$ & $u=x-\frac{1}{2}.~_{\square}$