Methods to solve differential equations We are given the equation
$$\frac{1}{f(x)} \cdot \frac{d\left(f(x)\right)}{dx} = x^3.$$ 
To solve it, "multiply by $dx$" and integrate:
$\frac{x^4}{4} + C = \ln \left( f(x) \right)$ 
But $dx$ is not a number, what does it mean when I multiply by $dx$, what am I doing, why does it work, and how can I solve it without multiplying by $dx$?  
Second question:
Suppose we have the equation $$\frac{d^2f(x)}{dx^2}=(x^2-1)f(x)$$
Then for large $x$, we have $\frac{d^2f(x)}{dx^2}\approx x^2f(x)$, with the approximate solution $ke \cdot \exp \left (\frac{x^2}{2} \right)$
Why is it then reasonable to suspect, or assume, that the solution to the original equation, will be of the form $f(x)=e^{x^2/2} \cdot g(x)$, where $g(x)$ has a simpler form then $f(x)$? When does it not work?  
Third question:
The method of replacing all occurences of $f(x)$, and its derivatives by a power series, $\sum a_nx^n$, for which equations does this work or lead to a simpler equation?
Do we lose any solutions this way?

FIRST QUESTION
This is answered by studying the fundamental theorem of calculus, which basically (in this context) says that if on an interval $(a,b)$
\begin{equation}
\frac{dF}{dx} = f
\end{equation}
then,
\begin{equation}
\int^{b}_{a} f\left(x\right) dx = F(b) - F(a)
\end{equation}
where the integral is the limit of the Riemann sum. And then, you can specify anti-derivatives (indefinite integrals) as $F\left(x\right) + c$ by fixing an arbitrary $F\left(a\right) = c$ and considering the function
\begin{equation}
F\left(x\right) = \int^{x}_{a} f\left(x\right) dx 
\end{equation}
where we do not specify the limits as they are understood to exist and are arbitrary. It is from here actually, that we get that multiplying by $dx$ rule.
For, 
\begin{equation}
F = \int dF
\end{equation}
and hence 
\begin{equation}
dF = f(x)dx \implies F = \int f(x) dx
\end{equation}
This is a little hand waiving, and for actual proofs, you would have to study Riemann sums.
Now, after this, for specific examples such as yours, I guess Andre Nicolas's answer is better but still I will try to offer something similar. 
We can say let $g\left(x\right) = \int x^{3} dx$ and $h\left(x\right) = log\left|f\left(x\right)\right|$. Then, 
\begin{equation}
\frac{dh\left(x\right)}{dx} = \frac{dg\left(x\right)}{dx} 
\end{equation}
for all $x \in \left(a,b\right)$
and hence, we can say that 
\begin{equation}
h\left(x\right)= g\left(x\right) + C
\end{equation}
