Solve ∫ left(3x+2right)^3 wrt x

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

Evaluate

$$\frac{\left(3x+2\right)^{4}}{12}+С$$

Solution Steps

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

$$\int 27x^{3}+54x^{2}+36x+8\mathrm{d}x$$

Integrate the sum term by term.

$$\int 27x^{3}\mathrm{d}x+\int 54x^{2}\mathrm{d}x+\int 36x\mathrm{d}x+\int 8\mathrm{d}x$$

Factor out the constant in each of the terms.

$$27\int x^{3}\mathrm{d}x+54\int x^{2}\mathrm{d}x+36\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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

$$\frac{27x^{4}}{4}+54\int x^{2}\mathrm{d}x+36\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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

$$\frac{27x^{4}}{4}+18x^{3}+36\int x\mathrm{d}x+\int 8\mathrm{d}x$$

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

$$\frac{27x^{4}}{4}+18x^{3}+18x^{2}+\int 8\mathrm{d}x$$

Find the integral of 8 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{27x^{4}}{4}+18x^{3}+18x^{2}+8x$$

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.

$$\frac{27x^{4}}{4}+18x^{3}+18x^{2}+8x+С$$

Differentiate w.r.t. x

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