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

Jul 6, 2015

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 $\mathrm{dx} = \mathrm{du}$

$= \int {u}^{5} {\left(u + 5\right)}^{2} \mathrm{du}$

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

$= \int {u}^{5} \left({u}^{2} + 10 u + 25\right) \mathrm{du}$

$= \int {u}^{7} + 10 {u}^{6} + 25 {u}^{5} \mathrm{du}$

$= {u}^{8} / 8 + \frac{10}{7} {u}^{7} + \frac{25}{6} {u}^{6}$

$= \textcolor{b l u e}{\frac{1}{8} {\left(x - 2\right)}^{8} + \frac{10}{7} {\left(x - 2\right)}^{7} + \frac{25}{6} {\left(x - 2\right)}^{6} + C}$

If one were to simplify this, eventually one would get:
$= \frac{1}{168} {\left(x - 2\right)}^{6} \left(21 {x}^{2} + 156 x + 304\right) + C$

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

(It gives $\frac{1}{8} {\left(x - 2\right)}^{8} + \frac{10}{7} {\left(x - 2\right)}^{7} + \frac{25}{6} {\left(x - 2\right)}^{6} - \frac{2432}{21} + C$, but

$\frac{2432}{21}$ IS a constant, which embeds into $C$)