Solve ∫ (2x-3)(5x+4) wrt x

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

Evaluate

$$\frac{10x^{3}}{3}-\frac{7x^{2}}{2}-12x+С$$

Solution Steps

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

$$\int 10x^{2}+8x-15x-12\mathrm{d}x$$

Combine 8x and -15x to get -7x.

$$\int 10x^{2}-7x-12\mathrm{d}x$$

Integrate the sum term by term.

$$\int 10x^{2}\mathrm{d}x+\int -7x\mathrm{d}x+\int -12\mathrm{d}x$$

Factor out the constant in each of the terms.

$$10\int x^{2}\mathrm{d}x-7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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

$$\frac{10x^{3}}{3}-7\int x\mathrm{d}x+\int -12\mathrm{d}x$$

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}.

$$\frac{10x^{3}}{3}-\frac{7x^{2}}{2}+\int -12\mathrm{d}x$$

Find the integral of -12 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{10x^{3}}{3}-\frac{7x^{2}}{2}-12x$$

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.

$$\frac{10x^{3}}{3}-\frac{7x^{2}}{2}-12x+С$$

Differentiate w.r.t. x

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