The problem is the Integrand of e^(4t)/(e^(2t)+3*e^(t)+2)
Any ideas on how to begin would be greatly appreciated.
Setting $\displaystyle e^{t}=x$ the function you have to integrate becomes...
$\displaystyle \frac {e^{4t}}{e^{2t}+3\cdot e^{t} + 2} = \frac{x^{4}}{x^{2} + 3\cdot x + 2} = \frac{x^{4}}{(1+x)\cdot (2+x)}$ (1)
Now is...
$\displaystyle \frac{1}{(1+x)\cdot (2+x)}= \frac{1}{1+x} -\frac{1}{2}\cdot \frac{1}{1+\frac{x}{2}}$ (2)
... and the two fractional terms in (2) can be written as series in the following manner...
$\displaystyle \frac{1}{1+x} = 1 - x + x^{2} - x^{3} + \dots$
$\displaystyle \frac{1}{1+\frac{x}{2}} = 1 - \frac {x}{2} + (\frac{x}{2})^{2} - (\frac{x}{2})^{3} + \dots$ (3)
Now you insert (3) in (2), then multiply by $\displaystyle x^{4}$ and obtain (1) , then set $\displaystyle x= e^{t}$ in (1) and [finally!...] at this point you can integrate 'term by term'... a little tedious job ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
A little p.s. : the series expansions (3) are possible only for $\displaystyle |x|<1$, so that the procedure is correct only for $\displaystyle |e^{t}|<1$, i.e. only for $\displaystyle t<0$...