Solve ∫ (3x-8)(2x+3) wrt x

$$\int{ (3x-8)(2x+3) }d x$$

Evaluate

$$2x^{3}-\frac{7x^{2}}{2}-24x+С$$

Solution Steps

Apply the distributive property by multiplying each term of 3x-8 by each term of 2x+3.

$$\int 6x^{2}+9x-16x-24\mathrm{d}x$$

Combine 9x and -16x to get -7x.

$$\int 6x^{2}-7x-24\mathrm{d}x$$

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int -7x\mathrm{d}x+\int -24\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-7\int x\mathrm{d}x+\int -24\mathrm{d}x$$

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

$$2x^{3}-7\int x\mathrm{d}x+\int -24\mathrm{d}x$$

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 -7 times \frac{x^{2}}{2}.

$$2x^{3}-\frac{7x^{2}}{2}+\int -24\mathrm{d}x$$

Find the integral of -24 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}-\frac{7x^{2}}{2}-24x$$

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.

$$2x^{3}-\frac{7x^{2}}{2}-24x+С$$

Differentiate w.r.t. x

$$\left(3x-8\right)\left(2x+3\right)$$