Question

find the indefinite integral and check your result by differentiation. (remember the constant of integration.)

$$ int ( x + 2 ) ( 3 x - 3 ) d x $$

Explanation:

Step1: Expand the integrand

$$\begin{align*} (x + 2)(3x-3)&=x(3x - 3)+2(3x - 3)\\ &=3x^{2}-3x + 6x-6\\ &=3x^{2}+3x - 6 \end{align*}$$

Step2: Integrate term - by - term

Use the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\))

$$\begin{align*} \int(3x^{2}+3x - 6)dx&=3\int x^{2}dx+3\int xdx-6\int dx\\ &=3\times\frac{x^{3}}{3}+3\times\frac{x^{2}}{2}-6x + C\\ &=x^{3}+\frac{3}{2}x^{2}-6x + C \end{align*}$$

Step3: Differentiate the result

Use the power rule for differentiation \((x^{n})^\prime=nx^{n - 1}\)

$$\begin{align*} \frac{d}{dx}(x^{3}+\frac{3}{2}x^{2}-6x + C)&=\frac{d}{dx}(x^{3})+\frac{3}{2}\frac{d}{dx}(x^{2})-6\frac{d}{dx}(x)+\frac{d}{dx}(C)\\ &=3x^{2}+3x-6 \end{align*}$$

And \(3x^{2}+3x - 6=(x + 2)(3x - 3)\)

Answer:

\(x^{3}+\frac{3}{2}x^{2}-6x + C\)