Question

How do you integrate `int ln(x+1)` by integration by parts method?

Answer 1

using `intln(x+1)dx=int(1xxln(x+1))dx`

and then putting
`u=ln(x+1)` `&` `(dv)/dx=1`

we end up with the result:

`I=(x+1)ln(x+1)-x+C`

Explanation:

Integration by parts formula:

`intu((dv)/dx)dx=uv-intv((du)/dx)dx`

write`intln(x+1)dx=int(1xxln(x+1))dx`

with intergation by parts it is usual to put `u=lnf(x)`

Let `u=ln(x+1)=>(du)/dx=1/(x+1)`

Let`(dv)/dx=1=>v=x`

`intu((dv)/dx)dx=xln(x+1)-int(x/(x+1))dx`

on spliting the improper fraction in the integral we have:

`intu((dv)/dx)dx=xln(x+1)-int(1-1/(x+1))dx`

integrating:

`intu((dv)/dx)dx=xln(x+1)-x+ln(x+1) +C`

which simplifies to:

`=(x+1)ln(x+1)-x+C`