Question

What is the integral of `xln^2(x+3)`?

Answer 1

I got

`(x^2 - 9)/2ln^2(x+3) - 1/2(x - 9)(x+3)ln(x+3) + 1/4(x+3)(x-21) + C`

---

DISCLAIMER: LONG ANSWER!

Whenever `ln` is in an integral, you should consider trying integration by parts to see where that gets you.

For `int udv`, what we have is:

`\mathbf(int udv = uv - intvdu)`

So the strategy here is to choose one term that is easier to differentiate, like `ln^2(x+3)`, and one term that is easier to antidifferentiate, like `x`.

FIRST INTEGRATION BY PARTS

Let:
`u = (ln(x+3))^2 -> du = (2ln(x+3))/(x+3)dx`
`dv = xdx -> v = x^2/2`

`=> x^2/2ln^2(x+3) - int (x^2ln(x+3))/(x+3)dx`

Now, you see how there's `ln(x+3)` and `1/(x+3)`? We can do another substitution. This is going to feel a bit pointless, but it'll give us a way out of this by allowing us to divide.

SOME U SUBSTITUTION

Let `u = x+3`. Then, `du = dx` and `x^2 = (u-3)^2 = u^2 - 6u + 9`.

So, what we now have is:

`=> x^2/2ln^2(x+3) - int ((u-3)^2lnu)/udu`

`=> x^2/2ln^2(x+3) - int ((u^2 - 6u + 9)lnu)/udu`

`=> x^2/2ln^2(x+3) - int (u^2lnu - 6u lnu + 9lnu)/udu`

`=> x^2/2ln^2(x+3) - int u lnu - 6lnu + (9lnu)/udu`

That's not too bad now.

SECOND INTEGRATION BY PARTS

The first integral is done by three different integration by parts, but fairly straightforward ones.

For `int sdt`, we have `\mathbf(int sdt = st - int tds)`. So...

For `int u lnudu`, let:
`s = lnu -> ds = 1/udu`
`dt = udu -> t = u^2/2`

`=> u^2/2 lnu - int u/2du = u^2/2 lnu - u^2/4`

THIRD INTEGRATION BY PARTS

For `6int lnudu`, let:
`s = lnu -> ds = 1/udu`
`dt = du -> t = u`

`=> u lnu - int u/udu = u lnu - u`

FOURTH INTEGRATION BY PARTS

For `9int lnu/udu`, let `s = lnu` and `ds = 1/udu`. Then:

`=> 9int sds = 9/2s^2 = 9/2ln^2u`

PUTTING IT ALL TOGETHER

So, for our overall integral, we currently have:

`=> x^2/2ln^2(x+3) - [int u lnudu - int6lnudu + int(9lnu)/udu]`

`= x^2/2ln^2(x+3) - [(u^2/2 lnu - u^2/4) - 6(u lnu - u) + 9/2ln^2u]`

Make sure you catch those parentheses!

`= x^2/2ln^2(x+3) - [u^2/2 lnu - u^2/4 - 6u lnu + 6u + 9/2ln^2u]`

`= x^2/2ln^2(x+3) - u^2/2 lnu + u^2/4 + 6u lnu - 6u - 9/2ln^2u`

But since `u = x+3`, we now have:

`= x^2/2ln^2(x+3) - (x+3)^2/2 ln(x+3) + (x+3)^2/4 + 6(x+3) ln(x+3) - 6(x+3) - 9/2ln^2(x+3)`

Regroup terms together:

`= x^2/2ln^2(x+3) - 9/2ln^2(x+3) + 6(x+3) ln(x+3) - (x+3)^2/2 ln(x+3) + (x+3)^2/4 - 6(x+3)`

`= (x^2/2 - 9/2)ln^2(x+3) + (-(x+3)^2/2 + 6x + 18)ln(x+3) + (x+3)^2/4 - 6(x+3) + C`

Yeah, I think I'll stop here.

If you wanted to work to simplify this some more, you can get this eventually through some expansion and re-factoring:

`= color(blue)((x^2 - 9)/2ln^2(x+3) - 1/2(x - 9)(x+3)ln(x+3) + 1/4(x+3)(x-21) + C)`

Turns out that this was right!