Why does a constant in an integral only sometimes have an antiderivate

Question

Summary

Consider the following indefinite integrals:

A) $$\int (2 + 3) dx = 5\ + C$$
B) $$\int (2 + 3x) dx = 2x + \frac{3x^2}{2}\ + C$$

As you can see, simply appending $x$ after $3$ also affects $2$. This doesn't appear to be consistent to me. I can see how A is correct assuming that you pull the constants out of the integration. I can also see how B is correct assuming you don't pull the constant out of the integral.

Question(s)

Why is the number $2$ treated differently between the two integrals? Or, as the title of the question implies: why aren't we finding the antiderivative of the number $2$ in A, but we ARE in B?

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Edit: 4/18/2018

As commenters have explained, it turns out that cymath is incorrectly evaluating the integral (A) above. It should be...

$$\int (2 + 3) dx = 5x\ + C$$

...which shows consistency between A and B.

Answer 1

The number $2$ isn't treated differently between the two antiderivatives.

Your first example $\int{2+3}dx$ can be rewritten as the following. $$\int{2x^0+3x^0}dx$$ Applying the same rule you used in your second example, it is easier to see that this becomes $5x+C$.

Your second example is correct. This antiderivative has an extra term so it gets dealt with accordingly. That is, $\int{2+3x}dx=2x+\frac{3x^2}{2}+C$.

In summary, the constant will always have an antiderivative. Reading the comments as I finish typing this answer, it appears that Cymath is incorrect in this case.

Answer 2

The reason is that the first formula contains a mistake, and hence is putting you on the wrong track: actually

$$\int (2 + 3)\ dx = (2 + 3)x + C = 5x + C$$

not $(2 + 3) + C$, or $5 + C$. The way to remember this is that you can always pull a constant out of an integral:

$$\int k\ dx = k \int dx$$

and the integral of $dx$ is simply $x$, up to a constant shift, because effectively you are just adding up all the small increments of $x$, which is what $dx$ represents, beginning with some unspecified starting value upon which we commence said incrementation.

With this mistake fixed, you should be able to see now the connection between this case and the case where we have a linear function inside the integral.