if i=integral (e^x) log(e^x+1)
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!!!