# How do you find the integral int_1^(3)12(1+5x)^5dx ?

Sep 7, 2014

There are a couple of ways, but the usual way to do this is by using the change-of-variable rule:
let $u = 1 + 5 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 5$,
then ${\int}_{1}^{3} \left(12 {\left(1 + 5 x\right)}^{5}\right) \mathrm{dx} = 12 {\int}_{1}^{3} {\left(1 + 5 x\right)}^{5} \cdot 5 \cdot \frac{1}{5} \mathrm{dx}$
$= 12 {\int}_{1}^{3} {u}^{5} \cdot \frac{1}{5} \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
$= \frac{12}{5} {\int}_{6}^{16} {u}^{5} \mathrm{du}$
and integrating normally from there.

However, if one isn't yet familiar with the change of variable rule, one can just as easily integrate this particular function rather easily (albeit slightly more prone to error) by expanding the brackets and integrating a particularly long function.

${\left(1 + 5 x\right)}^{5} = 3125 {x}^{5} + 3125 {x}^{4} + 1250 {x}^{3} + 250 {x}^{2} + 25 x + 1$ (this can be easily verified)

$\implies 12 {\int}_{1}^{3} {\left(1 + 5 x\right)}^{5} \mathrm{dx}$
$= 12 {\left[\frac{3125 {x}^{6}}{6} + \frac{3125 {x}^{5}}{5} + \frac{1250 {x}^{4}}{4} + \frac{250 {x}^{3}}{3} + \frac{25 x}{2} + x\right]}_{1}^{3}$

The plus side of the second method is that it's easy, and it can be used to check that you have your change-of-variable rule method correct.
The minus side is that it's more prone to errors, and takes up more time (which is valuable in tests and exams).