Solve ∫ -2x^3-3x^2+3 wrt x

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

Evaluate

$$-\frac{x^{4}}{2}-x^{3}+3x+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

Find the integral of 3 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

$$-\frac{x^{4}}{2}-x^{3}+3x+С$$

Differentiate w.r.t. x

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