Integration by substitution simplifying I wanna integrate $\int_{0}^{1} \frac{dx}{e^x-e^{-2x}+2}$but after substituting $e^x=t$ and simplifying I get trouble trying to use partial fraction expansion on the expression
$\frac{t}{(t+1)(t^2+t-1)}$. I know that $t^2+t-1=(t+\frac{1+\sqrt{5}}{2})(t+\frac{1-\sqrt{5}}{2})$ but this is troublesome to work with (not allowed to use calculators etc). Any suggestions to facilitate the calculations? 

You don’t need to factor the quadratic. Use the usual partial fractions technique:
$$\begin{align*}
\frac{t}{6^3+2t^2-1}&=\frac{A}{t+1}+\frac{Bt+C}{t^2+t-1}\\\\
&=\frac{(A+B)t^2+(A+B+C)t+(C-A)}{(t+1)(t^2+t-1)}\;,
\end{align*}$$
so $A+B=0=C-A$, and $A+B+C=1$. This is an easy system to solve: $C=A$, $B=-A$, and hence $A=1$, so we have
$$\int_1^e\left(\frac1{t+1}-\frac{t-1}{t^2+t-1}\right)dt\;,$$
assuming no silly errors on my part. The first term is easy; for the second, note that
$$\frac{t-1}{t^2+t-1}=\frac{(t+1)-2}{t^2+t-1}=\frac{t+1}{t^2+t-1}-\frac2{\left(t+\frac12\right)^2-\frac54}\;.$$
