Solve ∫ (3x+2)^3*(x)

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

Evaluate

$$\frac{27x^{5}}{5}+\frac{27x^{4}}{2}+12x^{3}+4x^{2}+С$$

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}.

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

Use the distributive property to multiply 27x^{3}+54x^{2}+36x+8 by x.

$$\int 27x^{4}+54x^{3}+36x^{2}+8x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 27x^{4}\mathrm{d}x+\int 54x^{3}\mathrm{d}x+\int 36x^{2}\mathrm{d}x+\int 8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$27\int x^{4}\mathrm{d}x+54\int x^{3}\mathrm{d}x+36\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 27 times \frac{x^{5}}{5}.

$$\frac{27x^{5}}{5}+54\int x^{3}\mathrm{d}x+36\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

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

$$\frac{27x^{5}}{5}+\frac{27x^{4}}{2}+36\int x^{2}\mathrm{d}x+8\int x\mathrm{d}x$$

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

$$\frac{27x^{5}}{5}+\frac{27x^{4}}{2}+12x^{3}+8\int x\mathrm{d}x$$

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

$$\frac{27x^{5}}{5}+\frac{27x^{4}}{2}+12x^{3}+4x^{2}$$

Simplify.

$$4x^{2}+12x^{3}+\frac{27x^{4}}{2}+\frac{27x^{5}}{5}$$

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.

$$4x^{2}+12x^{3}+\frac{27x^{4}}{2}+\frac{27x^{5}}{5}+С$$

Differentiate w.r.t. x

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