`int (x + 1)(3x - 2) dx` Find the indefinite integral.

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Chapter 4, 4.1 - Problem 23 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the indefinite integral, hence, you need to open the brackets such that:

`(x + 1)(3x - 2) = 3x^2 - 2x + 3x - 2 = 3x^2 + x - 2`

`int(x + 1)(3x - 2)dx = int (3x^2 + x - 2)dx`

You need to split the integral:

`int (3x^2 + x - 2)dx = int 3x^2 dx + int xdx - int 2dx`

You need to use the formula` int x^n dx = (x^(n+1))/(n+1) + c`

`int 3x^2 dx= 3x^3/3 + c => int 3x^2 dx= x^3 + c`

`int xdx = x^2/2 + c`

`int 2dx = 2x + c`

Gathering the results yields:

`int (3x^2 + x - 2)dx = x^3 + x^2/2 - 2x + c`

Hence, evaluating the indefinite integral, yields` int (3x^2 + x - 2)dx = x^3 + x^2/2 - 2x + c.`

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