Solve ∫ (5x^3+2x^2-6x+3) wrt x

$$\int{ (5 { x }^{ 3 } +2 { x }^{ 2 } -6x+3) }d x$$

Evaluate

$$\frac{5x^{4}}{4}+\frac{2x^{3}}{3}-3x^{2}+3x+С$$

Solution Steps

Integrate the sum term by term.

$$\int 5x^{3}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int -6x\mathrm{d}x+\int 3\mathrm{d}x$$

Factor out the constant in each of the terms.

$$5\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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 5 times \frac{x^{4}}{4}.

$$\frac{5x^{4}}{4}+2\int x^{2}\mathrm{d}x-6\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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

$$\frac{5x^{4}}{4}+\frac{2x^{3}}{3}-6\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -6 times \frac{x^{2}}{2}.

$$\frac{5x^{4}}{4}+\frac{2x^{3}}{3}-3x^{2}+\int 3\mathrm{d}x$$

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

$$\frac{5x^{4}}{4}+\frac{2x^{3}}{3}-3x^{2}+3x$$

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.

$$\frac{5x^{4}}{4}+\frac{2x^{3}}{3}-3x^{2}+3x+С$$

Differentiate w.r.t. x

$$5x^{3}+2x^{2}-6x+3$$