39. $int_{-1}^{1}(r + 1)^{2}dr$

Explanation:

Step1: Expand the integrand

$(r + 1)^2=r^{2}+2r + 1$ So, $\int_{1}^{1}(r + 1)^2dr=\int_{1}^{1}(r^{2}+2r + 1)dr$

Step2: Use the integral rules

$\int_{1}^{1}(r^{2}+2r + 1)dr=\left[\frac{r^{3}}{3}+r^{2}+r\right]_{1}^{1}$

Step3: Evaluate the definite - integral

$\left(\frac{1^{3}}{3}+1^{2}+1\right)-\left(\frac{1^{3}}{3}+1^{2}+1\right)$ $=\left(\frac{1}{3}+1 + 1\right)-\left(\frac{1}{3}+1 + 1\right)=0$

Answer:

$0$