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How do you integrate #(5x-3)^2dx#?

Answer 1

#int (5x-3)^2dx=1/15*(5x-3)^3+C#

Making the substitution #t=5x-3# then #dx=1/5dt# then we have #1/5*int t^2dt# using that #int x^ndx=x^(n+1)/(n+1)+C# with #n ne -1# then we obatin #int (5x-3)^2dx=1/15*(5x-3)^3+C#

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Answer 2

Axel Alvarez

To integrate ((5x - 3)^2 , dx), use the power rule for integration:

Expand the expression: ((5x - 3)^2 = 25x^2 - 30x + 9).
Integrate each term separately: (\int (25x^2 - 30x + 9) , dx).
Apply the power rule for integration:

[ \int x^n , dx = \frac{x^{n+1}}{n+1} + C ]

Integrate each term:

[ \int 25x^2 , dx = \frac{25x^3}{3} + C ] [ \int -30x , dx = -15x^2 + C ] [ \int 9 , dx = 9x + C ]

Combine the results:

[ \int (5x - 3)^2 , dx = \frac{25x^3}{3} - 15x^2 + 9x + C ]

Where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.