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

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Answer 1 · 2018-05-30T14:33:04

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

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 · 2018-05-30T14:34:46

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

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 · 2018-05-30T14:39:00

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

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

#(2x-5)(2x-5)#

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

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