I'd say that it is more a question of pattern recognition than anything. You may have to memorise a little bit to get going, but after then you should look at the integral and get an idea of whether you need a spanner or a screwdriver or a wrench.
I have long since forgotten what the integral of $\frac{1}{\sqrt{1-x^2}}$ is, but I remember that it looks $x=\sin\theta$-ish, and I try that. $\frac{x}{\sqrt{1-x^2}}$ has more of a $y=x^2$, $dy=2x\ dx$ feeling to it, though I may be wrong.
I've never memorised MathematicsStudent1122's $\int (\sin x)^n \ dx$, but I dare say that if I found myself having to confront it daily, I would remember it after the first few workings out.
Don't rot your brain with a calculator, because it will teach you nothing. People aren't asking you to work out integrals because they want to know the answer! If all else fails, work out a Taylor series, integrate term by term, and see if the answer looks familiar.
Due to criticism in the comments, I am limiting the scope of and rewriting
my answer. I still think I have something to offer even though I admit my answer is not complete.
The scope of the question itself is not entirely clear. There are possibly three questions: (1) How to integrate any function symbolically? (2) Does a "product rule" for integrals exist? (3) How do you integrate any rational function. I think (2) is a very interesting one.
I'll address (2) and (3) but (1) is outside of my knowledge. I am aware of related terms such as differential algebra and differential galois theory and the risch algorithm which may shed insight but, again, I am not knowledgeable in this.
For (2):
I suspect you are hoping for a formula such as the following. Let us take inspiration from the product rule for derivatives which has the form
$$\frac{d}{dx}(fg)=\lambda(f,f',g,g')$$
with $\lambda$ an elementary function defined as $\lambda(w,x,y,z)=wz+xy$. Let $F$ and $G$ be the antiderivatives of $f$ and $g$ respectively. Taking the differentiation case, you are hoping for a formula
$$\int fg dx = \lambda(f,F,g,G)$$
where $\lambda$ is some sort of function that's "nice". Implicit also is the "plus C". Specifically, I would say it is made of elementary functions or less precisely, the sort of functions you find in pre-calc. (I admit this does not cover all bases, however, the following will give at least some insight.)
Now take the case of $f=1/x$, $F=\log x$, $g=\sin x$ and $G = -\cos x$. We'd hope for some expression
$$\int \frac{\sin x}{x} dx = \lambda (1/x, \log x, \sin x, -\cos x).$$
However, it is well known that there is no closed form to the integral of the the so-called sinc function $\sin x/x$ in terms of elementary functions. But if $\lambda$ takes in elementary functions and combines them in an elementary way, the above expression would also have to be elementary. So no "nicely expressed", or elementary formula for an "integral product rule" exists.
This tells us at the least it would be a beast of a problem to find such a function. But in a way, this is backwards since the real work of showing $(\sin x)/x$ has no nice indefinite integral is itself a hard problem.
For (3):
Let me remind how you can deal with quadratic factors in partial fractions such as
$$r(x) = \sum \frac{p_i(x)}{q_i(x)}$$
where $q_i(x)$ is degree 2 and has complex roots and $p_i$ is degree 1 at most. Now you can always complete the square and add and subtract in the numerator to get a form
$$\int \frac{2c(x-a)+d}{(x-a)^2+b^2} dx = \int \frac{2c(x-a)}{(x-a)^2+b^2} + \frac{d}{(x-a)^2+b^2} dx.$$
The first term evaluates to $c\log((x-a)^2+x)$. The second to $d/b \arctan[(x-a)/b]$.
This is not a thorough explanation of how to compute the antiderivatives of rational functions but I want to make you aware of this point.
Best Answer
Too long for a comment.
I am going to give you another example of weird usage of terms but that we use it in this way because those were the term that stuck.
In probability, there exists the concept of renewal theory (events that once they happen, the entire process starts from scratch). A typical example is a return to the origin by the random walk. Once the random walk reaches the origin, you can forget the entire past and assume the random walk started from scratch. Now, in renewal theory there are two types of events that are of interest, those that can happen infinitely many times and those that will happen (almost) surely a finitely number of times. The latter type of event are usually called "transitory" which makes perfect intuitive sense. Unfortunately, the former are called "recurrent" which is not quite right since the expected time between two "recurrent" events can be infinite! (In other words, the amount of time you would expect to pass between two successive occurrences of this "recurrent" event is likely to be very large which makes the term "recurrent", literally "occuring often or frequently", a really bad choice of terminology.) The better term "persistent" event never become popular even though the events really are "persistent" and not "recurrent" (they can happen again and again even though they don't happen often). And since I am touching this topic, the "recurrent" events are further classified into two: those with finite waiting time and those with infinite waiting time; those with finite waiting time are really "recurrent" in the English sense.