Solve ∫ 5(x+2)-(x-1)(x+4)-6x

$$\int 5 ( x + 2 ) - ( x - 1 ) ( x + 4 ) - 6 x$$

Evaluate

$$-\frac{x^{3}}{3}-2x^{2}+14x+С$$

Solution Steps

Use the distributive property to multiply 5 by x+2.

$$\int 5x+10-\left(x-1\right)\left(x+4\right)-6x\mathrm{d}x$$

Apply the distributive property by multiplying each term of x-1 by each term of x+4.

$$\int 5x+10-\left(x^{2}+4x-x-4\right)-6x\mathrm{d}x$$

Combine 4x and -x to get 3x.

$$\int 5x+10-\left(x^{2}+3x-4\right)-6x\mathrm{d}x$$

To find the opposite of x^{2}+3x-4, find the opposite of each term.

$$\int 5x+10-x^{2}-3x-\left(-4\right)-6x\mathrm{d}x$$

The opposite of -4 is 4.

$$\int 5x+10-x^{2}-3x+4-6x\mathrm{d}x$$

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

$$\int 2x+10-x^{2}+4-6x\mathrm{d}x$$

Add 10 and 4 to get 14.

$$\int 2x+14-x^{2}-6x\mathrm{d}x$$

Combine 2x and -6x to get -4x.

$$\int -4x+14-x^{2}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -4x\mathrm{d}x+\int 14\mathrm{d}x+\int -x^{2}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-4\int x\mathrm{d}x+\int 14\mathrm{d}x-\int x^{2}\mathrm{d}x$$

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 -4 times \frac{x^{2}}{2}.

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

Find the integral of 14 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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 -1 times \frac{x^{3}}{3}.

$$-2x^{2}+14x-\frac{x^{3}}{3}$$

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.

$$-2x^{2}+14x-\frac{x^{3}}{3}+С$$

Differentiate w.r.t. x

$$14-4x-x^{2}$$