I don't do partial fractions till C4, this is a C3 question (AQA)
If you're going to use that substitution straight off, you're going to end up with partial fractions.
The only other thing I can suggest is to multiply top and bottom by e^(-x), and then use a similar substitution, which will get you round the partial fractions issue.
We've I=∫ex+12dx. Letting u=ex+1 gives ∫u(u−1)2du. You don't have to know the method of partial fractions to observe that 2=2u−2(u−1). Thus I=∫u(u−1)2u−2(u−1)du=2∫u−11du−2∫u1du=2ln(u−1)−2lnu+k=2lnex−2ln(ex+1)+k=2x−ln(ex+1)+k.