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.

$\require{begingroup}\begingroup$This should help to get closer to the final result (if you want to calculate this manually):
$$\newcommand{\dd}{\; \mathrm{d}} I=\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \dd x = 
\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x-\frac32)^2 -\frac14}} \dd x$$
It will take some computing but we can get that
$3x^3 - x^2 + 2x - 4 = 3(x-\frac32)^3+\frac{25}2(x-\frac32)^2+\frac{77}4(x-\frac32)+\frac{55}8$. 
So we have
$$I=
\int_{-3/2}^{-1/2} \frac{3u^3+\frac{25}2u^2+\frac{77}4u+\frac{55}8}{\sqrt{u^2-\frac14}} \dd u = 
\begin{vmatrix}
  t=2u & u=\frac t2 \\
  \dd t=2\dd u & \dd u = \frac12\dd t
\end{vmatrix} = 
\frac12 \int_{-3}^{-1} \frac{\frac{3t^3}8+\frac{25}8t^2+\frac{77}8t+\frac{55}8}{\sqrt{\frac{t^2}4-\frac14}} \dd t =
\frac18 \int_{-3}^{-1} \frac{3t^3+25t^2+77t+55}{\sqrt{t^2-1}} \dd t =
\frac18 \int_{-3}^{-1} \frac{3t(t^2-1)+25(t^2-1)+80t+80}{\sqrt{t^2-1}} \dd t = 
\frac18 \int_{-3}^{-1} (3t+25)\sqrt{t^2-1} +80 \frac{t+1}{\sqrt{t^2-1}} \dd t 
$$
You can check that Wolfram Alpha returns the same value for the original integral and this integral. (To be honest, I am not sure how I am supposed to get a PIN number from the result.)
Now you could divide up this into separate integrals which should be not too difficult:


*

*How to calculate this integral with square roots: $\int\frac{ \sqrt{x+1} }{ \sqrt{ x-1 }} \, dx$

*Indefinite integral of $\int\sqrt{x^2-1}dx$

*For $\int t\sqrt{t^2-1} \dd t$ the substitution $s=t^2-1$ seems reasonable.


$\endgroup$
