As André Nicolas says in his comment, it may be hard. I don't think one can get very far unless students have some intuition for polynomial long division, which, in my experience, is not always the case among calculus students.
If you can assume that your students understand the idea that, for polynomials $F(x)$ and $G(x),$
$$
\frac{F(x)}{G(x)}=Q(x)+\frac{R(x)}{G(x)}
$$
where polynomials $Q(x)$ and $R(x)$ are the quotient and remainder, the latter of which satisfies $\deg(R(x))<\deg(G(x)),$ then you can tell them that, since polynomials are straightforward to integrate, we only need to worry about integrating the second term, that is, a fraction in which the degree of the numerator is less than the degree of the denominator.
Dealing with the case where the denominator is a power of a binomial is then not so bad. In particular, we can answer your question about why there is no $x$ attached to the $B$ in your first example. The reason is that, if there were an $x$ term, it could be eliminated as in the following example:
$$
\frac{2x+7}{(x+3)^2}=\frac{2(x+3)+7-2\cdot3}{(x+3)^2}=\frac{2(x+3)+1}{(x+3)^2}=\frac{2}{x+3}+\frac{1}{(x+3)^2}.
$$
More generally, when the denominator is a binomial raised to a power greater than $2,$ all non-constant terms in the numerator can be eliminated:
$$
\begin{aligned}
\frac{4x^2+2x+7}{(x+3)^3}&=\frac{4(x+3)^2-4(6x+9)+2x+7}{(x+3)^3}=\frac{4(x+3)^2-22x-29}{(x+3)^3}\\
&=\frac{4(x+3)^2-22(x+3)+22\cdot3-29}{(x+3)^3}=\frac{4(x+3)^2-22(x+3)+37}{(x+3)^3}\\
&=\frac{4}{x+3}-\frac{22}{(x+3)^2}+\frac{37}{(x+3)^2}.
\end{aligned}
$$
Essentially we're doing a change of variable from $x$ to $u=x+3,$ and using the fact that a polynomial in $x$ is a polynomial in $u$ of the same degree, to get this conclusion.
The case where the denominator is the product of different binomials is best understood by first going in the opposite direction:
$$
\frac{5}{x+2}+\frac{4}{x+3}=\frac{5(x+3)+4(x+2)}{(x+2)(x+3)}=\frac{9x+23}{(x+2)(x+3)}.
$$
The method of partial fractions is then understood as the process of undoing this computation. This is always possible because the resulting pair of linear equations always has a unique solution. This is a consequence of the assumption that the two binomials in the denominator are different (more precisely, that they are not constant multiples of each other).
Students also need to understand that a quadratic denominator cannot always be factorized over the reals. (For example, $x^2+x+1.$) We therefore have to be able to deal with numerators up to degree $1.$ You can then show how fractions with degree $2$ and $1$ denominators get put together:
$$
\frac{2x+3}{x^2+x+1}+\frac{4}{x+2}=\frac{(2x+3)(x+2)+4(x^2+x+1)}{(x^2+x+1)(x+2)}=\frac{6x^2+11x+10}{(x^2+x+1)(x+2)}.
$$
Partial fractions is then the method for reversing this process. It again involves solving a system of linear equations which is guaranteed to have a unique solution (assuming the denominators have no common factor). Proving this involves algebra and linear algebra knowledge that is probably beyond most first year calculus students. But this example shows that, in general, the $x$ in the numerator of the last term in your first example is needed.
Another key result needed for complete understanding is that any polynomial with real coefficients can be factorized into degree $1$ and irreducible degree $2$ factors. To prove this requires knowledge that goes quite a bit beyond first-year calculus.
