Solve ∫ x^3-3x

$$\int x ^ { 3 } - 3 x$$

Evaluate

$$\frac{x^{4}}{4}-\frac{3x^{2}}{2}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

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

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

$$\frac{x^{4}}{4}-\frac{3x^{2}}{2}$$

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.

$$\frac{x^{4}}{4}-\frac{3x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x\left(x^{2}-3\right)$$