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:

HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
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.

Noticing that
$ 3 x^3-x^2+2 x-4=\frac{3}{2}(2 x-3)\left(x^2-3 x+2\right)+\frac{5}{2}\left(5 x^2-7 x+2\right),$
we have $$
I=\frac{3}{2} \underbrace{\int_0^1(2 x-3) \sqrt{x^2-3 x+2} d x}_J+\frac{5}{2} \underbrace{\int_0^1 \frac{5 x^2-7 x+2}{\sqrt{x^2-3 x+2}} d x}_K
$$$$
\begin{aligned}
J =\frac{3}{2} \int_0^1 \sqrt{x^2-3 x+2} d\left(x^2-3 x+2\right) =\left[\left(x^2-3 x+2\right)^{\frac{3}{2}}\right]_0^1=-2 \sqrt{2}
\end{aligned}
$$
Again, decompose the numerator of the integrand of $K$ into 3 parts as
$$
5 x^2-7 x+2=5\left(x^2-3 x+2\right)+4(2 x-3)+4,
$$
we get
$$
\begin{aligned}
K&=5  \int_0^1 \sqrt{x^2-3 x+2} d x+4 \int_0^1 \frac{d\left(x^2-3 x+2\right)}{\sqrt{x^2-3 x+2}} +4 \int \frac{d x}{\sqrt{x^2-3 x+2}}\\&= 5 \cdot \frac{1}{8}(6 \sqrt{2}-\ln (3+2 \sqrt{2}))-8 \sqrt{2}+4 \ln (3+2 \sqrt{2}) \cdots(*)\\&= -\frac{17}{4} \sqrt{2}+\frac{27}{8} \ln (3+2 \sqrt{2})
\end{aligned}
$$
where $(*)$ comes from my post.
Plugging back yields
$$
\boxed{I=\frac{135}{16} \ln (3+2 \sqrt{2})-\frac{101}{8} \sqrt{2}}
$$
