Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 34

Find the integral $\displaystyle \int^2_1 \frac{s^4 + 1}{s^2} ds$

Using 2nd Fundamental Theorem of Calculus

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$, where $F$ is any anti-derivative of $f$.

Let $\displaystyle f(s) = \frac{s^4 + 1 }{s^2}$ or $f(s) = s^2 + s^{-2}$, then


$
\begin{equation}
\begin{aligned}

F(s) =& \left( \frac{s^{2 + 1}}{2 + 1} \right) + \left( \frac{s^{-2 + 1}}{-2 + 1} \right) + C
\\
\\
F(s) =& \frac{s^3}{3} + \left( \frac{s^{-1}}{-1} \right) + C
\\
\\
F(s) =& \frac{s^3}{3} - s^{-1} + C
\\
\\
F(s) =& \frac{s^3}{3} - \frac{1}{s} + C

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \int^2_1 \frac{s^4 + 1}{s^2} ds = F(2) - F(1)
\\
\\
& \int^2_1 \frac{s^4 + 1}{s^2} ds = \frac{(2)^3}{3} - \frac{1}{2} + C- \left[ \frac{(1)^3}{3} - \frac{1}{1} + C \right]
\\
\\
& \int^2_1 \frac{s^4 + 1}{s^2} ds = \frac{8}{3} - \frac{1}{2} + C - \frac{1}{3} + 1 - C
\\
\\
& \int^2_1 \frac{s^4 + 1}{s^2} ds = \frac{7}{3} - \frac{1}{2} + 1
\\
\\
& \int^2_1 \frac{s^4 + 1}{s^2} ds = \frac{14 - 3 + 6}{6}
\\
\\
& \int^2_1 \frac{s^4 + 1}{s^2} ds = \frac{17}{6}


\end{aligned}
\end{equation}
$