Question

How do you use substitution to integrate `(x-2)^5 (x+3)^2 dx`?

Answer 1

You really only have two choices. `u = x-2` or `u = x+3`.

Let's use `u = x-2`. Thus:
`x + 3 - 5 + 5 = x - 2 + 5 = u + 5`

With `x + 3 = u + 5`,
`x = u + 2` and `dx = du`

`= int u^5(u+5)^2du`

(good, now we don't have to expand a 5th order term)

`= int u^5(u^2+10u+25)du`

`= int u^7 + 10u^6+25u^5du`

`= u^8/8 + 10/7u^7+25/6u^6`

`= color(blue)(1/8(x-2)^8 + 10/7(x-2)^7+25/6(x-2)^6 + C)`

If one were to simplify this, eventually one would get:
`= 1/168 (x-2)^6 (21x^2 + 156x + 304) + C`

Notice though that Wolfram Alpha would not agree with this answer, which is... odd.

(It gives `1/8(x-2)^8 + 10/7(x-2)^7+25/6(x-2)^6 - 2432/21 + C`, but

`2432/21` IS a constant, which embeds into `C`)