Solve ∫ (1+2x)^3 wrt x

$$\int ( 1 + 2 x ) ^ { 3 } d x$$

Evaluate

$$\frac{\left(2x+1\right)^{4}}{8}+С$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(1+2x\right)^{3}.

$$\int 1+6x+12x^{2}+8x^{3}\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int 6x\mathrm{d}x+\int 12x^{2}\mathrm{d}x+\int 8x^{3}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int 1\mathrm{d}x+6\int x\mathrm{d}x+12\int x^{2}\mathrm{d}x+8\int x^{3}\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x+6\int x\mathrm{d}x+12\int x^{2}\mathrm{d}x+8\int x^{3}\mathrm{d}x$$

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

$$x+3x^{2}+12\int x^{2}\mathrm{d}x+8\int x^{3}\mathrm{d}x$$

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

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

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

$$x+3x^{2}+4x^{3}+2x^{4}$$

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.

$$x+3x^{2}+4x^{3}+2x^{4}+С$$

Differentiate w.r.t. x

$$\left(2x+1\right)^{3}$$