Verified answer

The simplest way is to multiply it out

(3x-2)^2 = 9x^2 - 12x + 4

Integrate to obtain 3x^3 - 6x^2 + 4x + C.

You can also use a substitution u=3x-2 and du=3dx and dx=du/3

INTEGRAL[(1/3)u^2 du] = (1/9)u^3 +C = (1/9)(3x-2)^3 + C

This can be verified by multiplying it out and comparing it to the original answer. All antiderivatives vary by a constant.

Integral Of 3x 2

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RE:

What is the integral of (3x-2)^2?

I know that inside the parenthesis is (3x^2/2 -2x) but I don't know what to do with the square outside the parenthesis.

integral (-2+3 x)^2 dx:

For the integrand (3 x-2)^2, substitute u = 3 x-2 and du = 3 dx:

1/3 integral u^2 du

The integral of u^2 is u^3/3:

u^3/9+constant

Substitute back for u = 3 x-2:

1/9 (3 x-2)^3+constant

Which is equivalent for restricted x values to:

3 x^3-6 x^2+4 x+constant

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lower bound = 3x^2 ??? If so, answer = -6x sin(9x^4 + 1)

∫ (3x - 2)² dx

Note that you can expand (3x - 2)² to get:

∫ (3x - 2)(3x - 2) dx

==> ∫ 9x² - 6x - 6x + 4 dx

==> ∫ 9x² - 12x + 4 dx

==> ∫ 9x² dx - ∫ 12x dx + ∫ 4 dx [Set each term with ∫.]

==> 9 ∫ x² dx - 12 ∫ x dx + 4 ∫ 1 dx [Factor out the constant]

Then, by ∫ xⁿ dx = x^(n + 1)/(n + 1) + c:

9x^(2 + 1)/(2 + 1) - 12x^(1 + 1)/(1 + 1) + 4x + c

==> 9x³/3 + 12x²/2 + 4x + c

==> 3x³ + 6x² + 4x + c

I hope this helps!

use u substitution

u=3x-2

du=3dx

du/3=dx

so

int(u)^2*du/3

then when you integrate

(1/9)(3x-2)^3 + c

i = ∫ 9 x ² - 6x + 4 dx

I = 3 x ³ - 3 x ² + 4x + C

[(3x-2)^3]/9