About a step in the analysis of Quicksort by Sedgewick and Wayne In the book Algorithms, 4th Edition by Robert Sedgewick and Kevin Wayne, when they are analyzing quicksort (page 294), they present the sequence of transformations: 
$$\begin{gather*}
C_N = N + 1 + (C_0 + C_1 + \dots + C_{N-2} + C_{N-1})/N + (C_{N-1} + C_{N-2} + \dots + C_0)/N\\
NC_N = N(N+1) + 2(C_0 + C_1 + \dots + C_{N-2} + C_{N-1})\\
NC_N - (N-1)C_{N-1} = 2N + 2C_{N-1}\\
C_N/(N+1) = C_{N-1}/N + 2/(N+1)\\
C_N\sim 2(N+1)(1/3 + 1/4 + \dots + 1/(N+1))\end{gather*}$$

How did they get the last transformation? 

It is also written that the parenthesized quantity in the last expression  is the discrete estimate of the area under the curve $2/x$ from $3$ to $N$? How is it related to quicksort?
Let $\rho_N = C_N/(N+1)$. The next to last equation shows that
$$
\rho_N = \rho_{N-1} + \frac{2}{N+1}.
$$
Therefore
$$
\rho_N = \frac{2}{N+1} + \cdots + \frac{2}{2+1} + \rho_1.
$$
Multiplying by $N+1$,
$$
C_N = (N+1) \left(\frac{2}{3} + \cdots + \frac{2}{N+1} + \rho_1\right).
$$
Since $\rho_1$ is a constant whereas $2/3 + \cdots + 2/(N+1) \to \infty$, we deduce that
$$
C_N \sim (N+1) \left(\frac{2}{3} + \cdots + \frac{2}{N+1}\right).
$$
We can estimate this expression in many ways, for example by approximating the series by an integral (this is what the authors suggest when they mention the area under the curve $2/x$). Or we can recognize that it is roughly equal to $2H_{N+1}$, twice the $(N+1)$st harmonic number, and so using the well-known estimate $H_N \sim \log N$,
$$
C_N \sim 2N\log N.
$$

Finally, regarding the relevance to Quicksort, my guess is that $C_N$ is the average number of comparisons performed on a random array of length $N$; but you should be able to tell by reading the book.