Solve ∫ (x+9)^3 wrt x

$$\int ( x + 9 ) ^ { 3 } d x$$

Evaluate

$$\frac{\left(x+9\right)^{4}}{4}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+9\right)^{3}.

$$\int x^{3}+27x^{2}+243x+729\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 27x^{2}\mathrm{d}x+\int 243x\mathrm{d}x+\int 729\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+27\int x^{2}\mathrm{d}x+243\int x\mathrm{d}x+\int 729\mathrm{d}x$$

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}.

$$\frac{x^{4}}{4}+27\int x^{2}\mathrm{d}x+243\int x\mathrm{d}x+\int 729\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+9x^{3}+243\int x\mathrm{d}x+\int 729\mathrm{d}x$$

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

$$\frac{x^{4}}{4}+9x^{3}+\frac{243x^{2}}{2}+\int 729\mathrm{d}x$$

Find the integral of 729 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+9x^{3}+\frac{243x^{2}}{2}+729x$$

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.

$$\frac{x^{4}}{4}+9x^{3}+\frac{243x^{2}}{2}+729x+С$$

Differentiate w.r.t. x

$$\left(x+9\right)^{3}$$