Solve ∫ (x+1)(2x-5) wrt x

$$\int ( x + 1 ) ( 2 x - 5 ) d x$$

Evaluate

$$\frac{2x^{3}}{3}-\frac{3x^{2}}{2}-5x+С$$

Solution Steps

Apply the distributive property by multiplying each term of x+1 by each term of 2x-5.

$$\int 2x^{2}-5x+2x-5\mathrm{d}x$$

Combine -5x and 2x to get -3x.

$$\int 2x^{2}-3x-5\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{2}\mathrm{d}x+\int -3x\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{2}\mathrm{d}x-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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

$$\frac{2x^{3}}{3}-3\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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 -3 times \frac{x^{2}}{2}.

$$\frac{2x^{3}}{3}-\frac{3x^{2}}{2}+\int -5\mathrm{d}x$$

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{2x^{3}}{3}-\frac{3x^{2}}{2}-5x$$

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.

$$\frac{2x^{3}}{3}-\frac{3x^{2}}{2}-5x+С$$

Differentiate w.r.t. x

$$\left(2x-5\right)\left(x+1\right)$$