Answer
$$\int(\frac{1+r}{r})^2dr=-\frac{1}{r}+2\ln|r|+r+C$$
Work Step by Step
$$A=\int(\frac{1+r}{r})^2dr$$ $$A=\int(\frac{1+2r+r^2}{r^2})dr$$ $$A=\int(\frac{1}{r^2}+\frac{2}{r}+1)dr$$ $$A=\int(r^{-2}+\frac{2}{r}+1)dr$$ From Table 1, $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ $$\int cf(x)dx=c\int f(x)dx$$ Therefore, $$A=\int(r^{-2})dr+2\int(\frac{1}{r})dr+\int1dr$$
From Table 1, we also get the followings $$\int (x^n)dx=\frac{x^{n+1}}{n+1}+C(n\ne-1)$$ $$\int(\frac{1}{x})dx=\ln |x|+C$$ $$\int kdx=kx+C$$
Therefore, $$A=\frac{r^{-1}}{-1}+2\ln|r|+r+C$$ $$A=-\frac{1}{r}+2\ln|r|+r+C$$
