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?
 A: We need to use the factors of the denominator of the following integral, to decompose into partial fractions.  We need for the numerator of the second degree factor (second fraction below) to be of the form of a degree one polynomial: $Bx + C$, since $x^2 + 2$ cannot be factored into the product of first-degree factors:
$$\int \frac{3x^2 - 4x + 5}{(x-1)(x^2+2)}\,dx = \int \frac{A}{x- 1} + \frac{Bx + C}{x^2 + 2}\,dx\tag{1}$$
Then we need for $$A(x^2 + 2) + (Bx + C)(x-1) = 3x^2 - 4x + 5\tag{2}$$
We expand the terms on the left-hand side of equation $(2)$, group coefficients for $\,x^2,\, x,\; \text{and constants},\;$ and equate with the right-hand side of $(2)$:
$$\begin{align} A(x^2 + 2) + (Bx + C)(x-1) & = (A x^2 + 2A) + (Bx^2 + Cx - Bx - C)\\ \\ & = \color{blue}{\bf (A + B)}x^2 + \color{red}{\bf (C - B)} x + \color{green}{\bf (2A - C)}  \\ \\ & = \color{blue}{\bf 3}x^2 + \color{red}{\bf - 4}x + \color{green}{\bf 5}\end{align}$$
Now we match up coefficients of the left hand side with those on the right (I'll use color to distinguish between the three matchings we'll need to make), to obtain three equations in three unknowns $A,\, B, \,C :$: 
This gives us the following system of linear equations to solve:
$$\begin{align} \color{blue}{\bf A + B} & = \color{blue}{\bf 3}\\ 
\color{red}{\bf C - B} & = \color{red}{\bf -4} \\ 
\color{green}{\bf 2A - C} & = \color{green}{\bf 5} \end{align}$$
suppose we add the first two of these equations. Then we get: $$\begin{align} A + B - B + C & = 3 - 4 \\ A + C & = -1\end{align}$$. Now adding this "new" equation with the third equation above, we get $$\begin{align} 2A - C + A + C & = 5 - 1\\ 3A & = 4 \\ A & = \dfrac 43\end{align}$$
Now, we know that $2A -C = 5,$ so $$\begin{align} 2A - C & = 5 \\ C & = 2A - 5 \\
C & = 2\cdot \dfrac 43 -\dfrac{15}{3} \\
C &= \left(-\dfrac 73\right)
\end{align}.$$
And we know that $$\begin{align} A + B & = 3 \\ B & = 3 - A = 3 - \dfrac 43 \\ B& = \dfrac 53\end{align}$$
That gives us, in all: $$A = \dfrac 43,\;B = \frac 53,\;C =  -\dfrac 73$$
Now, we have the integral:
$$\begin{align} \int \frac{A}{x- 1} + \frac{Bx + C}{x^2 + 2}\,dx & =  \int \frac 13 \left(\frac{4}{x - 1} + \frac{5x - 7}{x^2 + 2}\right)\,dx \\
& = \frac 43 \int \dfrac {dx}{x - 1}  + \frac {1\cdot 5}{3\cdot 2} \int \frac{2x \,dx}{x^2 + 2} - \frac 73 \int \frac {dx}{x^2 + (\sqrt 2)^2} \\
& = \frac 43 \ln|x - 1| + \frac 56 \ln(x^2 + 2) - \frac 7{3\sqrt 2} \tan^{-1}\left(\frac{x}{\sqrt 2}\right) + C
\end{align}$$
We can get really clever and write the sum (first two terms) $ \dfrac 43\ln|x - 1| + \dfrac 56\ln(x^2 + 2)$ as $\ln(x - 1)^{4/3} + \ln(x^2 + 2)^{5/6} = \ln\left((x-1)^{4/3}(x^2 + 2)^{5/6}\right) = \ln\left((x - 1)^8(x^2 + 2)^5\right)^{1/6}$, giving us an answer: 
$$\ln\left((x - 1)^8(x^2 + 2)^5\right)^{1/6} - \frac 7{3\sqrt 2} \tan^{-1}\left(\frac{x}{\sqrt 2}\right) + C = $$
A: 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.
A: Indeterminate coefficients is a very general technique in mathematics that consists of three main steps.
Step 1: Find a general form (depending on parameters) that you can be sure it includes the solution of your problem.
Step 2: Impose the general form to be the solution of your problem and by reducing to some canonical form deduce that this corresponds to a system of equations in the parameters. In this problem of partial fraction decomposition what is meant by canonical form is to write the polynomials as sums of powers of $x$, which is a canonical form for polynomials.
Step 3: Solve the system of equations. 
Step 1 is what ensures that there is going to be at least a solution for Step 3. 
For example: Suppose we have the rational function
$$\frac{2}{(x-1)(x+1)}.$$
It is easy to see that $$\frac{2}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}$$
and this is its partial fraction decomposition. But assume we didn't know that and we try to find a partial fraction decomposition in the form $$\frac{2}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}.$$
This is equivalent to asking that $$2(x-1)=A(x-1)(x+1)+B(x+1).$$
Writing both sides as sums of powers of $X$ (this what Step 2 is about) we get
$$2x-2=Ax^2+Bx+(-A+B).$$
This gives us too many equations for only $2$ unknowns. Namely,
\begin{align*}
A&=0\\
B&=2\\
-A+B&=-2
\end{align*}
This system doesn't have solution. This means only one thing: That the fraction $2/(x-1)(x+1)$ cannot be written in the form $A/(x-1)+B/(x-1)^2$, i.e. this general form with parameters $A$ and $B$ doesn't contain the partial fraction decomposition of the given fraction, and therefore was not a good choice for Step 1. That is why in Step 1 you really need to make sure the general form contains the solution of your problem.
In the statement of your problem Step 1 is done incorrectly. The same phenomenon is happening. 
Moral: Indeterminate coefficient does require you to memorize something. What is the general form of the decomposition to search for.
How should it be?
How to prescribe a general enough partial fraction decomposition to be able to ensure there is going to be a solution?
Step A. Make sure the numerator has degree smaller than the denominator (it is a proper fraction). This is important, otherwise what comes after may not work. If this is not the case use long division to write your fraction as a polynomial plus a proper fraction.
Step B. Factorize the denominator in factors of the form $(x-r)^n$ and $(x^2+px+q)^m$ for different $r$, $p$, $q$'s, and there $p^2-4q<0$.
Step C. Prescribe for each factor $(x-r)^n$ a sum of the form
$$\frac{A_1}{(x-r)}+\frac{A_2}{(x-r)^2}+\ldots\frac{A_n}{(x-r)^n},$$
where there are $n$ summands, the same as the exponent $n$ in the factorization of the denominator. Likewise, for every $(x^2+px+q)^m$ prescribe a sum of the form 
$$\frac{A_1x+B_1}{(x^2+px+q)}+\frac{A_2x+B_2}{(x^2+px+q)^2}+\ldots\frac{A_mx+B_m}{(x^2+px+q)^m}.$$
It is a no-so-difficult exercise in algebra to show that this will ensure there is a solution to the indeterminate coefficient problem and a unique one.
Example: In your problem we have the rational function $\frac{3x^2-4x+5}{(x-1)(x^2+2)}$ you have already the denominator factored in the form required. Therefore you can search for a decomposition of the form $\frac{A}{(x-1)}+\frac{Bx+C}{x^2+2}$ and find a solution for the coefficients.
A: Before moving to partial fractions, it's useful to separate the denominator's derivative, which is $(3x^2-2x+2)$:
$$\begin{align}
\int\frac{3x^2-4x+5}{(x-1)(x^2+2)}\,dx
&=\int\frac{(3x^2-2x+2)+(-2x+3)}{(x-1)(x^2+2)}\,dx\\
&=\ln((x-1)(x^2+2))+\int\frac{-2x+3}{(x-1)(x^2+2)}\,dx\\
\end{align}$$
Now the remaining separation into partial fractions is easier because there's an extra zero in the system of equations. In fact the separation must be in the form:
$$\begin{align}
\ln((x-1)(x^2+2))+\int\left(\frac{A}{x-1}+\frac{-Ax+B}{x^2+2}\right)\,dx\\
\end{align}$$
to make the $x^2$ terms disappear from the numerator. Now the system to solve is $$\left\{\begin{array}{rcl}A+B&=&-2\\2A-B&=&3\end{array}\right.$$ "Adding" equations reveals that $A=1/3$. From there, we deduce $B=-7/3$. So we have $$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{-x+7}{x^2+2}\right)\,dx\\
\end{align}$$
Again, I'd separate a derivative:
$$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)+\frac{1}{3}\int\frac{-x+7}{x^2+2}\,dx\\
=\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\int\frac{2x-14}{x^2+2}\,dx\\
=\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{x^2+2}\,dx\\
\end{align}$$ and the last integral can be handled via substituting $2u^2=x^2$:
$$\begin{align}
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{x^2+2}\,dx\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{1}{3}\int\frac{7}{2u^2+2}\,(\sqrt{2}\,du)\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{7}{3\sqrt{2}}\arctan(u)+C\\
\ln((x-1)(x^2+2))+\frac{1}{3}\ln(x-1)-\frac{1}{6}\ln(x^2+2)+\frac{7}{3\sqrt{2}}\arctan(x/\sqrt{2})+C\\
\end{align}$$
