Integral $\int {t+ 1\over t^2 + t - 1}dt$ 
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$

Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} dt + B\int {1 \over t+w_2}dt \\=  {w - 1\over w - w_2} \ln|t + w| + {1- w_2\over w - w_2}\ln|t + w_2| + C $$
After finding the value of $w, w_2$ final answer I got is 
$${\sqrt{5} + 1\over 2\sqrt{5}}\ln|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\ln|t + 1/2 + \sqrt{5}/2| + C$$
But the given answer is :
$$\bbox[7px,Border:2px solid black]{ \frac{\ln\left(\left|t^2+t-1\right|\right)}{2}+\frac{\ln\left(\left|2t-\sqrt{5}+1\right|\right)-\ln\left(\left|2t+\sqrt{5}+1\right|\right)}{2\cdot\sqrt{5}}+C}$$
Where did I go wrong ? especially that first term of the answer is a mystery to me.

Note that we have
$$\begin{align}
& {\sqrt{5} + 1\over 2\sqrt{5}}\log|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\log|t + 1/2 + \sqrt{5}/2| + C\\[10pt]
&=  \frac12\log\left(|t + 1/2 - \sqrt{5}/2|\,|t + 1/2 + \sqrt{5}/2|\right)\\[10pt]
&{}+\frac{1}{2\sqrt{5}}\left(\log\left(|t + 1/2 - \sqrt{5}/2|\right)-\log\left(|t + 1/2 + \sqrt{5}/2|\right)\right)+C\\\\
&=\frac12\log\left(|t^2+t+1|\right)+\frac{1}{2\sqrt{5}}\left(\log\left(|t + 1/2 - \sqrt{5}/2|\right)-\log\left(|t + 1/2 + \sqrt{5}/2|\right)\right)+C\\\\
\end{align}$$
