Solve ∫ (6x^2-4x-2) wrt x

$$\int{ (6 { x }^{ 2 } -4x-2) }d x$$

Evaluate

$$2x^{3}-2x^{2}-2x+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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}.

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

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$2x^{3}-2x^{2}-2x$$

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.

$$2x^{3}-2x^{2}-2x+С$$

Differentiate w.r.t. x

$$2\left(x-1\right)\left(3x+1\right)$$