Find the slope and area under the graph for a given function

Let

$f(x) = \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \qquad \text{for} \quad x> 0.$

Determine the slope of the graph of $f$ at the point with $x$ -coordinate 1.
Find the volume of the solid of revolution formed by rotating the region between the graph of $f(x)$ and the interval $[1,4]$ about the $x$ -axis.

To take this derivative, using logarithmic differentiation will be easier,
$\begin{align*} \log (f(x)) &= \log \left( \sqrt{ \frac{4x+2}{x(x+1)(x+2)}} \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log (x(x+1)(x+2)) \right) \\[9pt] &= \frac{1}{2} \left( \log (4x+2) - \log x - \log (x+1) - \log (x+2) \right). \end{align*}$

Then differentiating both sides we have,

$\begin{align*} &&\frac{f'(x)}{f(x)} &= \frac{1}{2} \left( \frac{4}{4x+2} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right) \\[9pt] \implies && f'(x) &= \frac{1}{2} \sqrt{\frac{4x+2}{x(x+1)(x+2)}} \left( \frac{2}{2x+1} - \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} \right). \end{align*}$

So, to find the slope at the point with $x = 1$ we evaluate,

$f'(1) = \frac{1}{2} \left( \frac{2}{3} - 1 - \frac{1}{2} - \frac{1}{3} \right) = -\frac{7}{12}.$
First, the integral to compute the volume of the solid of revolution is,
$\begin{align*} V &= \pi \int_1^4 (f(x))^2 \, dx \\[9pt] &= \int_1^4 \frac{\pi(4x+2)}{x(x+1)(x+2)} \, dx. \end{align*}$

To evaluate this we use the partial fraction decomposition,

$\frac{2x+1}{x(x+1)(x+2)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2}.$

This gives us the equation

$A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) = 2x+1.$

Evaluating at $x = 0$ , $x = -1$ , and $x = -2$ we obtain

$A = \frac{1}{2}, \quad B = 1, \quad C = -\frac{3}{2}.$

Therefore, we have

$\begin{align*} V &= 2 \pi \int_1^4 \left( \frac{1}{2x} + \frac{1}{x+1} - \frac{3}{2(x+2)} \right) \, dx \\[9pt] &= 2 \pi \left( \frac{1}{2} \int_1^4 \frac{1}{x} \,dx + \int_1^4 \frac{1}{x+1} \, dx - \frac{3}{2} \int_1^4 \frac{1}{x+2} \, dx \right) \\[9pt] &= \pi \log x \Bigr \rvert_1^4 + 2 \pi \log |x+1| \Bigr \rvert_1^4 - 3 \pi \log |x+2| \Bigr \rvert_1^4 \\[9pt] &= \pi \log 4 + 2 \pi (\log 5 - \log 2) - 3 \pi (\log 6 - \log 3) \\[9pt] &= 2 \pi \log 2 + 2 \pi \log 5 - 2 \pi \log 2 - 3 \pi \log (2 \cdot 3) + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 - 3 \pi \log 3 + 3 \pi \log 3 \\[9pt] &= 2 \pi \log 5 - 3 \pi \log 2 \\[9pt] & = \pi (\log 25 - \log 8) \\[9pt] & = \pi \log \frac{25}{8}. \end{align*}$