Solve ∫ (x+3)(x-3)^2 wrt x

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

Evaluate

$$\frac{x^{4}}{4}-x^{3}-\frac{9x^{2}}{2}+27x+С$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$\int \left(x+3\right)\left(x^{2}-6x+9\right)\mathrm{d}x$$

Use the distributive property to multiply x+3 by x^{2}-6x+9 and combine like terms.

$$\int x^{3}-3x^{2}-9x+27\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int -3x^{2}\mathrm{d}x+\int -9x\mathrm{d}x+\int 27\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x-3\int x^{2}\mathrm{d}x-9\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-3\int x^{2}\mathrm{d}x-9\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}-x^{3}-9\int x\mathrm{d}x+\int 27\mathrm{d}x$$

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 -9 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{4}-x^{3}-\frac{9x^{2}}{2}+\int 27\mathrm{d}x$$

Find the integral of 27 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}-x^{3}-\frac{9x^{2}}{2}+27x$$

Simplify.

$$-\frac{9x^{2}}{2}-x^{3}+\frac{x^{4}}{4}+27x$$

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.

$$-\frac{9x^{2}}{2}-x^{3}+\frac{x^{4}}{4}+27x+С$$

Differentiate w.r.t. x

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