Solve ∫ (x+1)(2x+2)(x+1)

$$\int ( x + 1 ) ( 2 x + 2 ) ( x + 1 )$$

Evaluate

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

Solution Steps

Multiply x+1 and x+1 to get \left(x+1\right)^{2}.

$$\int \left(x+1\right)^{2}\left(2x+2\right)\mathrm{d}x$$

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

$$\int \left(x^{2}+2x+1\right)\left(2x+2\right)\mathrm{d}x$$

Apply the distributive property by multiplying each term of x^{2}+2x+1 by each term of 2x+2.

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

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$\int 2x^{3}+6x^{2}+4x+2x+2\mathrm{d}x$$

Combine 4x and 2x to get 6x.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

$$\frac{x^{4}}{2}+6\int x^{2}\mathrm{d}x+6\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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

$$\frac{x^{4}}{2}+2x^{3}+6\int x\mathrm{d}x+\int 2\mathrm{d}x$$

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}.

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

Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

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.

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

Differentiate w.r.t. x

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