It's not as bad as the other, but it ain't pretty neither.
You could use a clever partial fraction decomp on it, may be one way.
I actually believe that by some really farfetched coincidence my answer is right.
The denominator can factor as:
You can then use partial fractions lol
Using and , let and , then the integral becomes
Now I'm going to collapse and sleep for a while.
One question: Krizalid's solution will not work if the domain of integration includes 0 (or does it?). Why does DivideBy0's solution work for such a region?
The reason for the "almost" is that there are a couple of typos(?) in DivideBy0's answer. Eg 2x + 1 rather than etx. .....