# How do you integrate  (2x-5)^2 dx?

May 30, 2018

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx} = {\left(2 x - 5\right)}^{3} / 6 + C$

#### Explanation:

Substitute:

$t = \left(2 x - 5\right)$

$\mathrm{dt} = 2 \mathrm{dx}$

so:

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx} = \frac{1}{2} \int {t}^{2} \mathrm{dt} = {t}^{3} / 6 + C$

Undoing the substitution:

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx} = {\left(2 x - 5\right)}^{3} / 6 + C$

We can do the same thing using a simpler notation:

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx} = \frac{1}{2} \int {\left(2 x - 5\right)}^{2} d \left(2 x - 5\right) = {\left(2 x - 5\right)}^{3} / 6 + C$

May 30, 2018

$= \frac{4 {x}^{3}}{3} - 10 {x}^{2} + 25 x + c$

#### Explanation:

Easiest way to do this is to expand ${\left(2 x - 5\right)}^{2}$:

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx}$

$\int \left(2 x - 5\right) \left(2 x - 5\right) \mathrm{dx}$

$\int \left(4 {x}^{2} - 20 x + 25\right) \mathrm{dx}$

Then integrate each term separately and leave out the signs to make things easier:

$\int \left(4 {x}^{2}\right) \mathrm{dx} - \int \left(20 x\right) \mathrm{dx} + \int \left(25\right) \mathrm{dx}$

$\int \left(4 {x}^{2}\right) \mathrm{dx} = \frac{4 {x}^{3}}{3}$

$\int \left(20 x\right) \mathrm{dx} = \frac{20 {x}^{2}}{2} = 10 {x}^{2}$

$\int \left(25\right) \mathrm{dx} = 25 x$

Therefore,

$\int \left(4 {x}^{2} - 20 x + 25\right) \mathrm{dx} = \frac{4 {x}^{3}}{3} - 10 {x}^{2} + 25 x$

Do not forget constant as this is an indefinite integral:

$\int \left(4 {x}^{2} - 20 x + 25\right) \mathrm{dx} = \frac{4 {x}^{3}}{3} - 10 {x}^{2} + 25 x + c$

May 30, 2018

$= \frac{4 {x}^{3}}{3} - 10 {x}^{2} + 25 x + C$

#### Explanation:

$\int {\left(2 x - 5\right)}^{2} \mathrm{dx}$

$\left(2 x - 5\right) \left(2 x - 5\right)$

$4 {x}^{2} - 10 x - 10 x + 25$

$\frac{4 {x}^{2 + 1}}{3} \frac{- 20 {x}^{2}}{2} + 25 x + C$