Solve ∫ (5-2x)(4x+3) wrt x

$$\int ( 5 - 2 x ) ( 4 x + 3 ) d x$$

Evaluate

$$-\frac{8x^{3}}{3}+7x^{2}+15x+С$$

Solution Steps

Apply the distributive property by multiplying each term of 5-2x by each term of 4x+3.

$$\int 20x+15-8x^{2}-6x\mathrm{d}x$$

Combine 20x and -6x to get 14x.

$$\int 14x+15-8x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 14x\mathrm{d}x+\int 15\mathrm{d}x+\int -8x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$14\int x\mathrm{d}x+\int 15\mathrm{d}x-8\int x^{2}\mathrm{d}x$$

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

$$7x^{2}+\int 15\mathrm{d}x-8\int x^{2}\mathrm{d}x$$

Find the integral of 15 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$7x^{2}+15x-8\int x^{2}\mathrm{d}x$$

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 -8 times \frac{x^{3}}{3}.

$$7x^{2}+15x-\frac{8x^{3}}{3}$$

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.

$$7x^{2}+15x-\frac{8x^{3}}{3}+С$$

Differentiate w.r.t. x

$$\left(-4x-3\right)\left(2x-5\right)$$