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

You can try a hyperbolic substitution: $$\sqrt{x^2-3x+2}=\sqrt{\Bigl(x-\frac{3\strut}2\Bigr)^{2}-\frac94+2}=\frac12\sqrt{(2x-3)^2-1},$$ so you can set, for $t\ge 0$, $$2x-3=\cosh t\iff x=\frac{\cosh t+3}2,\qquad\mathrm dx=\frac12\sinh t\,\mathrm dt.$$ and the denominator of the integrand becomes $\;\frac12\sinh t$.

After you've simplified, you should obtain the integral of a cubic polynomial in $\cosh t$.

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 $$\int{\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}dx={{x}^{2}}H\left( x \right)}+\frac{13}{4}xH\left( x \right)+\frac{101}{16}H\left( x \right)+\frac{135}{16}\int{\frac{dx}{H\left( x \right)}}$$