Solve ∫ (3x-1)(x+3)

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

Evaluate

$$x^{3}+4x^{2}-3x+С$$

Solution Steps

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

$$\int 3x^{2}+9x-x-3\mathrm{d}x$$

Combine 9x and -x to get 8x.

$$\int 3x^{2}+8x-3\mathrm{d}x$$

Integrate the sum term by term.

$$\int 3x^{2}\mathrm{d}x+\int 8x\mathrm{d}x+\int -3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$3\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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}. Multiply 3 times \frac{x^{3}}{3}.

$$x^{3}+8\int x\mathrm{d}x+\int -3\mathrm{d}x$$

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

$$x^{3}+4x^{2}+\int -3\mathrm{d}x$$

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

$$x^{3}+4x^{2}-3x$$

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.

$$x^{3}+4x^{2}-3x+С$$

Differentiate w.r.t. x

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