Trying to figure out how this integration happened, can someone explain this to me? 
We have $$\frac{f’(x)}{f(x)} = -5,$$
  so that
  $$\int \frac{f’(x)}{f(x)}\,dx = \int (-5)\,dx.$$
  This implies that $\left|\ln(f(x))\right| = -5x+C$, so that $f(x) = Ke^{-5x}$, where $C$ and $K$ are arbitrary constants. Setting $f(0)=2$ implies that $K=2$. Thus, the answer is (d).

This is the integration. Wouldn't the integral of $f'(x)$ become $f(x)$? How does it just disappear in the answer? 
 A: If I properly understand what you're asking about, note the integral is of the expression $\frac{f'(x)}{f(x)}$, not just $f'(x)$, so you can't do an integral of just that numerator by itself but, instead, you need to do it of the entire fraction together.
The solution works as it does because, using that $\frac{d(\ln(x))}{dx} = \frac{1}{x}$ along with the chain rule, you get
$$\frac{d(\ln(f(x))}{dx} = \frac{f'(x)}{f(x)} \tag{1}\label{eq1A}$$
In the given solution, the Fundamental Theorem of Calculus was also used to get the reverse, i.e., that
$$\int \frac{f'(x)}{f(x)}dx = \ln(\left|f(x)\right|) + C \tag{2}\label{eq2A}$$
for some constant $C$. In the solution you're asking about, the $C$ appears on the right side instead after integrating $-5$ to get $-5x + C$.
A: You want to understand how the integral $$\int\frac{f'(x)}{f(x)}\mathrm dx$$ was evaluated. Well, make the substitution $f(x)=y$ to get $$\int\frac1y\mathrm dy,$$ which evaluates to $\log |y|,$ minus the arbitrary constant.
A: *

*The answer of ${\it John\ Omielan}$ is basically OK, except both in his answer and your solution have a typo:
It shall be $\ln|f(x)|$ instead of $|\ln(f(x))|$, whenever the absolute value appears.

*Another typo in the posted solution is that: $K$ is not arbitrary constant. It shall be:
$K$ is an arbitrary nonzero constant.
In general, if you don't have the information $f(0)=2$, the correct solution of the differential equation is a series of function $f(x) = Ke^{-5x}$, where $K\neq 0$ is an arbitrary constant.
