Question

\[\int\limits {u \over 1+u} du\]

Answer 1

take 1+u=v
then u=v-1.. solve

Answer 2

i think thats its long division..

Answer 3

yes

Answer 4

\[Original~ question~~~ \int\limits {e^x \over 1+e^x}dx--->x=Inu\]

Answer 5

Once you do long div, you'll get
\[\text{}=\int\limits \left(1-\frac{1}{x+1}\right) \, dx\]

Answer 6

u=e^x+1and \[du=e^xd x\]
\[\int\limits \frac{1}{u} \, du\]