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? 

Since 
$$\frac{t}{(t + 1)(t^2 + t - 1)} = \frac{1}{t + 1} - \frac{t - 1}{t^2 + t - 1} = \frac{1}{t + 1} -\frac{1}{2}\frac{2t + 1}{t^2 + t - 1} + \frac{3}{2}\frac{1}{t^2 + t - 1},$$
you may integrate each term on the right separately. The first one will give you $\log|t + 1| + C$. The second one will also give you a log: use the $u$-substitution $u = t^2 + t - 1$ and note that $du = (2t + 1)\, dt$. Even the last one will give you a log. Write $t^2 + t - 1 = (t + 1/4)^2 - 5/4$ and note that by partial fraction decomposition, 
$$\int \frac{dv}{v^2 - a^2} = \frac{1}{2}\log\left|\frac{v - a}{v + a}\right| + C.$$
