Solve ∫ (x-3)(x+2) wrt x=

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

Evaluate

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x+С$$

Solution Steps

Apply the distributive property by multiplying each term of x-3 by each term of x+2.

$$\int x^{2}+2x-3x-6\mathrm{d}x$$

Combine 2x and -3x to get -x.

$$\int x^{2}-x-6\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -6\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{3}}{3}-\int x\mathrm{d}x+\int -6\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -1 times \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+\int -6\mathrm{d}x$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}-6x+С$$

Differentiate w.r.t. x

$$\left(x-3\right)\left(x+2\right)$$