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

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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 $C$ )

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