Solve ∫ (x-3)(x+1)

$$\int ( x - 3 ) ( x + 1 )$$

Evaluate

$$\frac{x^{3}}{3}-x^{2}-3x+С$$

Solution Steps

Apply the distributive property by multiplying each term of x-3 by each term of x+1.

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

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

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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 -2 times \frac{x^{2}}{2}.

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

Find the integral of -3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$\frac{x^{3}}{3}-x^{2}-3x+С$$

Differentiate w.r.t. x

$$\left(x-3\right)\left(x+1\right)$$