Both methods are correct. Your mistake is in the third line, you just made a mistake in your integration.
With this correction, your other 2x vanishes and you are left with a similar expression.
@ Gralla55:
There's nothing wrong with your solutions.
Also it's not necessary to use 'absolute value signs' because e^(-x)+1>0 for each x.
@DivisionByZero:
That's not a mistake because:
2ln(e^x+1)-2x=2ln(e^x+1)-2ln(e^x)=2ln((e^x+1)/(e^x))=2ln(e^(-x)+1)
So just the same as Gralla55 said.