Evaluating $\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$ I've got one integration question which I first felt was not a hard nut to crack. But, as I proceeded, difficulties arose. This is the one:

$\displaystyle\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$

I went ahead simplifying the two expressions and ultimately I reached this step:
$\displaystyle\int_{0}^{1}{\sqrt{\frac{x-1}{x-2}} \ (3x^2 + 2x + 4) \ dx}$
I don't now what to do now, had I followed the correct pathway? Is there any other simpler method? 

It can be seen easily that for $H\left( x \right)=\sqrt{{{x}^{2}}-3x+2}$ 
$$\begin{align}
  & {{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}=\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}, \\ 
 & {{\left( xH\left( x \right) \right)}^{\prime }}=\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}, \\ 
 & {{\left( H\left( x \right) \right)}^{\prime }}=\frac{2x-3}{2H\left( x \right)} \\ 
\end{align}$$
This suggests that there exists $a,b,c\ and\ d$
$$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a{{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}+b{{\left( xH\left( x \right) \right)}^{\prime }}+c{{\left( H\left( x \right) \right)}^{\prime }}+\frac{d}{H\left( x \right)}$$
Hence
$$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}+b\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}+c\frac{2x-3}{2H\left( x \right)}+\frac{d}{H\left( x \right)}$$ 
Or
$$3{{x}^{3}}-{{x}^{2}}+2x-4=\frac{a}{2}\left( 6{{x}^{3}}-15{{x}^{2}}+8x \right)+\frac{b}{2}\left( 4{{x}^{2}}-9x+4 \right)+\frac{c}{2}\left( 2x-3 \right)+d$$
Equating the coefficients yields 
$$\begin{align}
  & a=1 \\ 
 & -15a+4b=-2 \\ 
 & 8a-9b+2c=4 \\ 
 & 4b-3c+2d=-8 \\ 
\end{align}$$
and we get
$$a=1,b=\frac{13}{4},c=\frac{101}{16},d=\frac{135}{16}$$
Finally 
$$\int{\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}dx={{x}^{2}}H\left( x \right)}+\frac{13}{4}xH\left( x \right)+\frac{101}{16}H\left( x \right)+\frac{135}{16}\int{\frac{dx}{H\left( x \right)}}$$
