Solve ∫ 2x(x-3) wrt x

$$\int 2 x ( x - 3 ) d x$$

Evaluate

$$\frac{2x^{3}}{3}-3x^{2}+С$$

Solution Steps

Use the distributive property to multiply 2x by x-3.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

$$\frac{2x^{3}}{3}-6\int x\mathrm{d}x$$

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 -6 times \frac{x^{2}}{2}.

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

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.

$$\frac{2x^{3}}{3}-3x^{2}+С$$

Differentiate w.r.t. x

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