Question

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

Answer 1

`int (2x-5)^2 dx = (2x-5)^3/6+C`

Explanation:

Substitute:

`t = (2x-5)`

`dt = 2dx`

so:

`int (2x-5)^2 dx = 1/2 int t^2dt = t^3/6+C`

Undoing the substitution:

`int (2x-5)^2 dx = (2x-5)^3/6+C`

We can do the same thing using a simpler notation:

`int (2x-5)^2 dx = 1/2 int (2x-5)^2 d(2x-5) = (2x-5)^3/6+C`

Answer 2

`= (4x^3)/3 - 10x^2 + 25x + c`

Explanation:

Easiest way to do this is to expand `(2x-5)^2`:

`int (2x-5)^2 dx`

`int (2x-5)(2x-5) dx`

`int (4x^2-20x+25) dx`

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

`int (4x^2) dx - int (20x) dx + int (25) dx`

`int (4x^2) dx = (4x^3)/3`

`int (20x) dx = (20x^2)/2 = 10x^2`

`int (25) dx = 25x`

Therefore,

`int (4x^2-20x+25) dx = (4x^3)/3 - 10x^2 + 25x`

Do not forget constant as this is an indefinite integral:

`int (4x^2-20x+25) dx = (4x^3)/3 - 10x^2 + 25x + c`

Answer 3

`= (4x^3)/3 -10x^2 +25x + C`

Explanation:

`int (2x-5)^2 dx`

`(2x-5)(2x-5)`

`4x^2 -10x -10x +25`

`(4x^(2+1))/3 (-20x^2)/2 +25x + C`