Question

What is the antiderivative of `ln(x+1)`?

Answer 1

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

Explanation:

We have

`I=intln(x+1)dx`

We will use integration by parts:

`intudv=uv-intvdu`

For `intln(x+1)dx`, let `u=ln(x+1)` and `dv=dx`. These imply that `du=1/(x+1)dx` and `v=x`.

Thus,

`I=xln(x+1)-intx/(x+1)dx`

Solving the remaining integral:

`intx/(x+1)dx=int(x+1-1)/(x+1)dx=int1-1/(x+1)dx`

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

So,

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

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