Solve ∫ (r+1)^2dr

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

Evaluate

$$\frac{\left(r+1\right)^{3}}{3}+С$$

Solution Steps

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

$$\int r^{2}+2r+1\mathrm{d}r$$

Integrate the sum term by term.

$$\int r^{2}\mathrm{d}r+\int 2r\mathrm{d}r+\int 1\mathrm{d}r$$

Factor out the constant in each of the terms.

$$\int r^{2}\mathrm{d}r+2\int r\mathrm{d}r+\int 1\mathrm{d}r$$

Since \int r^{k}\mathrm{d}r=\frac{r^{k+1}}{k+1} for k\neq -1, replace \int r^{2}\mathrm{d}r with \frac{r^{3}}{3}.

$$\frac{r^{3}}{3}+2\int r\mathrm{d}r+\int 1\mathrm{d}r$$

Since \int r^{k}\mathrm{d}r=\frac{r^{k+1}}{k+1} for k\neq -1, replace \int r\mathrm{d}r with \frac{r^{2}}{2}. Multiply 2 times \frac{r^{2}}{2}.

$$\frac{r^{3}}{3}+r^{2}+\int 1\mathrm{d}r$$

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

$$\frac{r^{3}}{3}+r^{2}+r$$

If F\left(r\right) is an antiderivative of f\left(r\right), then the set of all antiderivatives of f\left(r\right) is given by F\left(r\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{r^{3}}{3}+r^{2}+r+С$$

Differentiate w.r.t. r

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