Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 28

Find the integrals $\displaystyle \int^2_1 \frac{y+5y^2}{y^3} dy$

$\begin{equation} \begin{aligned} \int \frac{y+5y^2}{y^3} dy &= \int \left( \frac{y}{y^3} + \frac{5y^7}{y^3} \right) dy\\ \\ \int \frac{y+5y^2}{y^3} dy &= \int \left( \frac{1}{y^2} + 5y^4 \right) dy\\ \\ \int \frac{y+5y^2}{y^3} dy &= \int y^{-2} dy + 5 \int y^4 dy\\ \\ \int \frac{y+5y^2}{y^3} dy &= \left( \frac{y^{2+1}}{2+1} \right) + 5 \left( \frac{y^{4+1}}{4+1} \right) + C\\ \\ \int \frac{y+5y^2}{y^3} dy &= \frac{y^{-1}}{-1} + \cancel{5} \left( \frac{y^5}{\cancel{5}} \right) + C\\ \\ \int \frac{y+5y^2}{y^3} dy &= -y^{-1} + y^5 + C\\ \\ \int \frac{y+5y^2}{y^3} dy &= \frac{-1}{y} + y^5 + C\\ \\ \int^2_1 \frac{y+5y^2}{y^3} dy &= \frac{-1}{2} + (2)^5 + C - \left[ \frac{-1}{1} + (1)^5 + C \right]\\ \\ \int^2_1 \frac{y+5y^2}{y^3} dy &= \frac{-1}{2} + 32 + C + 1 - 1 - C\\ \\ \int^2_1 \frac{y+5y^2}{y^3} dy &= \frac{-1+64}{2}\\ \\ \int^2_1 \frac{y+5y^2}{y^3} dy &= \frac{63}{2} \end{aligned} \end{equation}$