<s>[INST] Partial fractions $\int \frac{(3x^2 - 4x + 5)\,dx}{(x-1)(x^2+2)}$ $$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)}$$
I am going to use undetermined coefficients since it seems straightforward, no wacky matrices or tables to memorize.
 $$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)} \quad = \quad\int \left(\frac {A}{x} + \frac{B}{(x-1)^2}+\frac{C}{(x-1)}+D\right)\,dx$$
I get $$A(x^3 - 3x^2 + 3x - 1) +B(x^3 - 2x^2 + x) + C(x^2 - x) + Dx$$
This gives me the sets of
$$\begin{align} 
A + B & = 0\\ \\ 
-3A + -2B + C & = 0 \\ \\
3A + B - C + D & = 0 \\ \\
-A &= 0
\end{align}$$
This is obviously wrong because A and B are now 0. What do I do?
 [/INST] When you factor $2x^2-3$ you get $2(x-\sqrt {\frac 32})(x+\sqrt{\frac 32})$ so you want $\frac 1{2x^2-3}=\frac A{x-\sqrt {\frac 32}}+\frac B{x+\sqrt {\frac 32}}$
Added:  note that this applies to a previous version of the question.
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