Question

How do you integrate `xln(1+x) dx`?

Answer 1

The solution is shown here (sometimes Wolfram Alpha has trouble computing integrals with `lnx` properly, so I gave the backwards check).

I would start with a u-substitution and separate the integral.

Let:
`u = x+1`
`du = dx`
`x = u - 1`

`=> int (u-1)lnudu`

`= int u lnudu - int lnudu`

With these two integrals in mind, we can do Integration by Parts (assuming you already know the integral of `lnx`). Ignoring the `int lnu`, let:

`s = lnu`
`ds = 1/udu`
`dt = udu`
`t = u^2/2`

Thus:

`st - int tds`

`= [(u^2lnu)/2 - int u^2/2*1/udu] - int lnudu`

`= (u^2lnu)/2 - 1/2int udu - int lnudu`

`= (u^2lnu)/2 - u^2/4 - (u lnu - u)`

`= (u^2lnu)/2 - u^2/4 - u lnu + u`

`= (u^2lnu)/2 - u lnu - u^2/4 + u`

`= ((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + 1 + C`

And `1` gets embedded into `C`:

`= color(blue)(((x+1)^2ln(x+1))/2 - (x+1) ln(x+1) - (x+1)^2/4 + x + C)`