Solve ∫ (x-3)(x^2-6x+2) wrt x

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

Evaluate

$$\frac{x^{4}}{4}-3x^{3}+10x^{2}-6x+С$$

Solution Steps

Use the distributive property to multiply x-3 by x^{2}-6x+2 and combine like terms.

$$\int x^{3}-9x^{2}+20x-6\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -9x^{2}\mathrm{d}x+\int 20x\mathrm{d}x+\int -6\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-9\int x^{2}\mathrm{d}x+20\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-9\int x^{2}\mathrm{d}x+20\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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 -9 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}-3x^{3}+20\int x\mathrm{d}x+\int -6\mathrm{d}x$$

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

$$\frac{x^{4}}{4}-3x^{3}+10x^{2}+\int -6\mathrm{d}x$$

Find the integral of -6 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-3x^{3}+10x^{2}-6x$$

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.

$$\frac{x^{4}}{4}-3x^{3}+10x^{2}-6x+С$$

Differentiate w.r.t. x

$$\left(x-3\right)\left(x^{2}-6x+2\right)$$