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

3 Answers
May 30, 2018

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

May 30, 2018

= (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

May 30, 2018

= (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