Solve ∫ (x+1)(x^2+2) wrt x

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

Evaluate

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

Solution Steps

Use the distributive property to multiply x+1 by x^{2}+2.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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

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

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}.

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

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

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

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.

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

Differentiate w.r.t. x

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