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

Question

1672 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-06T05:53:47

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#)