Partial fractions decomposition. Why $cx+d$ instead of $cx$ for the numerator of $(x^2+2$). I understand that the aim of partial fractions decomp. is simply to reach (an) integrable functions, but then I have trouble wrapping my head around why you cannot make the numerator of something like $x^2+2$ equal to $Cx$ alone and then later use u-sub. The question I tried this on and failed was this:enter image description here


In my opinion, the best way to understand why this is the case is to consider the partial fraction decomposition over $\mathbb{C}$ instead. It is much more intuitive once you get it. Here is the theorem from Fundamentals of Complex Analysis by Saff and Snider p. 105:
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If
$$R_{m,n}(z) = \frac{a_0 + a_1z + a_2z^2 + \cdots + a_mz^m}{b_n(z-\zeta_1)^{d_1}(z-\zeta_2)^{d_2} \cdots (z- \zeta_r)^{d_r}}$$
is a rational function whose denominator degree $n = d_1 + d_2 + \cdots + d_r$ exceeds its numerator degree $m$, then $R_{m,n}(z)$ has a partial fraction decomposition of the form
$$R_{m,n}(z) = \frac{A_0^{(1)}}{(z-\zeta_1)^{d_1}} + \frac{A_1^{(1)}}{(z- \zeta_1)^{d_1 - 1}} + \cdots + \frac{A_{d_1-1}^{(1)}}{(z-\zeta_1)} \\
+ \frac{A_0^{(2)}}{(z-\zeta_2)^{d_2}} + \cdots + \frac{A_{d_2 - 1}^{(2)}}{(z - \zeta_2)}\\
+ \cdots
+ \frac{A_0^{(r)}}{(z- \zeta_r)^{d_r}} + \cdots + \frac{A_{d_r - 1}^{(r)}}{(z- \zeta_r)}$$
where the $\lbrace A_s^{(j)}$ are constants. (The $\zeta_k$'s are assumed distinct.)
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Really try to understand how the subscript of $A$ behaves. With your example this becomes that
$$\frac{x^3 +5}{x^2(x^2+2)} = \frac{x^3 + 5}{x^2(x + i \sqrt{2})(x - i \sqrt{2})} = \frac{A_0^{(1)}}{x^2} + \frac{A_1^{(1)}}{x} + \frac{A_0^{(2)}}{x + i \sqrt{2}} + \frac{A_0^{(3)}}{x - i \sqrt{2}}.$$
Now, by the Fundamental Theorem of Algebra, any quadratic in the denominator will necessarily have two complex roots yielding the above behaviour. We can put all of these together and have a linear term ($bX + c$ form where $b$ and $c$ are some combination of the $A_s^{(j)}$'s) on the top.
So why look at it this way? Well, because then this implies that one can deal with partial fraction decomposition in the same way and even generalize the previous observation to understand the following intuitively. If you have a polynomial of degree $n$ on the denominator then its partial fraction decomposition will have a term which such a polynomial in the denominator, and in which the numerator is a polynomial with degree at most $n-1$. This for $n = 2$ is precisely your problem.
