A basic question about voltage current relation for an inductor
Eugene's comment is a compact statement of just how I was thinking when I read your question. His writing of it "sings" in my mind like a melody I know well. But it probably could use some elaboration for you.

Perhaps it is simpler to start with this simple statement in algebra:

$$I_L=\frac{V_L}{L}\cdot t$$

The above would assume that at the start of the period of time the current in the inductor was zero. It also assumes that either \$V_L\$ is a constant, or else that it is the effective average value if it varied somehow during the indicated period of time. But I think you can see that this statement strongly states that it requires time. Zero time would mean zero current, here. So at its root you can see that it takes time. But this does not discuss the question you ask, which is about "sudden" changes. This is where calculus (which is no more than algebra, modified to add a new kind of special variable that can only hold infinitely small, but still non-zero, values -- those special variables noted by placing \$\textrm{d}\$ in front of them.)

The above expression is more exactly written this way (there are still more exact ways of writing it, that I'll avoid here):

$$\textrm{d}I_L=\frac{V_{L}}{L}\cdot \textrm{d}t$$

It's still the same thing as before except that I've replaced the finite variable \$t\$, which can only hold finite values, with the infinitesimal variable \$\textrm{d}t\$, which can only hold infinitesimal but non-zero values. (Obviously, the factor \$\frac{V_L}{L}\$ remains a ratio of finite values.) The product of a finite ratio, times an infinitesimally small value, must itself be infinitesimally small. So the left side reflects this fact by replacing the earlier finite variable with its infinitesimal equivalent.

But let's look at this latest equation, now. What does it say? It says that in order to make even the smallest, most infinitesimally small change in the current of the inductor, you must apply a non-zero voltage for at least an infinitesimally small period of time.

In short, it takes time (and a non-zero applied voltage) to make a change in the current. I don't even need the integral to say so.

So why is the integral added? Because infinitesimally small values are pretty much useless for engineers making stuff. They need finite values, not infinitely small ones. So how many of these tiny values are needed to make a finite value? An infinite number of them, of course. So the special integral sign was invented.

Keeping in mind that whatever you do to one side of an equation you have to also do to the other side:

$$\begin{align*}
\textrm{d}I_L&amp;=\frac{V_{L}}{L}\cdot \textrm{d}t\\\\
\int \textrm{d}I_L&amp;=\int \frac{V_{L}}{L}\cdot \textrm{d}t\\\\
I_L&amp;=\int \frac{V_{L}}{L}\cdot \textrm{d}t
\end{align*}$$

The last step, looking at the left side change, is true because if you sum of an infinite number of infinitely small pieces of X, you get X. Obviously. If you divide up anything into an infinite number of tiny bits, then pasting all those bits back together must be what you originally started out with. So this last step is a "no brainer" in calculus.

Since \$L\$ is often taken as constant, then it should be obvious that I can treat \$\frac{V_{L_1}}{L}\cdot \textrm{d}t+\frac{V_{L_2}}{L}\cdot \textrm{d}t+\frac{V_{L_3}}{L}\cdot \textrm{d}t+ ...\$ by extracting the \$\frac{1}{L}\$ part and performing the sum without it, first, applying that fraction later. So:

$$\begin{align*}
I_L&amp;=\frac{1}{L}\int V_{L}\cdot \textrm{d}t
\end{align*}$$

A person familiar with calculus will just see the above immediately as the same as:

$$\textrm{d}I_L=\frac{V_{L}}{L}\cdot \textrm{d}t$$

So when the writer wrote as they did, they were seeing the core I started with above -- which is that to make a change in the current you must make a change in the time. So "it takes time for current to change." So from the integral, the differential obviously follows, and from the differential the conclusion then follows. That's how one gets there from here, I suppose.

You asked:


  What is that to do with integration that sudden change is not
  possible?


So in a very meaningful sense, it has nothing at all to do with the integration. Instead, it has to do with the differential equation that results when you mentally strip the integral away from it.



I'm sorry for elaborating so much on this. But I want to point out that there is a thinking process that you really need to struggle with yourself to acquire. It is crucial, as it influences your very ability to actually "see." You are blind before and have vision after. It's that much of "night and day." And it changes the way you see everything around you.