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}
$