Solve ∫ x^3-4x^2+3x wrt x

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

Evaluate

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

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -4 times \frac{x^{3}}{3}.

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

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}.

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

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.

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

Differentiate w.r.t. x

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