How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point. 

The question was:

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PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
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Anyone knows? Are we gonna get an integer number?


My attempt:
Does this help?

$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
  (long division)
  \begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
  after the substitution
  \begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}

Update: This may help us proceed.

I tried to proceed:

$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
  after $u^2+1=t$ and $dt=2udu$
  \begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}

I wonder whether this question can be solved from here?

Update: 
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.

Alternative method:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
Express the integrand in the form $\lfrac{(2ax+b)·(x^2-3x+2)+(2cx+d)}{\sqrt{x^2-3x+2}}$ for some constants $a,b,c,d$.
Then split it into $( a(2x-3) + (3a+b) ) · \sqrt{x^2-3x+2} + \lfrac{c(2x-3)+(3c+d)}{\sqrt{x^2-3x+2}}$, so that as a sum of four terms the first and third have obvious antiderivatives. The other two terms can be solved by standard techniques.
