Okh to solve your question I will use

i for integration and d for diffrentiation w.r.t. x okh

Apply ILATE rule

which says i= log(e^x+1) i e^xdx- i { d (log(e^x+1)) i e^xdx}dx

log(e^x+1)e^x- i {e^x/(e^x+1)*e^x}dx ( i e^x=e^x && d logx=1/x)

Now use substitution

[Let u=e^x then du=e^xdx]

The equation becomes

log(e^x+1)e^x- i {u/1+u}du

log(e^x+1)e^x- i {u+1-1/1+u}du

log(e^x+1)e^x- [ i du - i (1/1+u)du]

log(e^x+1)e^x- [u -log(1+u)] [ i ( 1/1+x)dx=log(1+x)]

Put u=e^x

log(e^x+1)e^x- [e^x -log(1+e^x)]

Hope this helps!!!