Calculate $\int_3^4 \sqrt {x^2-3x+2} \, dx$ using Euler's substitution

Using the third substitution of Euler

$$\sqrt{{{x}^{2}}-3x+2}=\sqrt{\left( x-1 \right)\left( x-2 \right)}=\left( x-1 \right)t$$ $$x=\frac{2-{{t}^{2}}}{1-{{t}^{2}}}\Rightarrow dx=\frac{2t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}dt$$ $$\begin{align} & =-\int_{1/\sqrt{2}}^{\sqrt{2/3}}{\frac{2{{t}^{2}}}{{{\left( {{t}^{2}}-1 \right)}^{3}}}dt} \\ & =\left[ \frac{\left( t+{{t}^{3}} \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}+\ln \left( \frac{1-t}{1+t} \right) \right]_{1/\sqrt{2}}^{\sqrt{2/3}} \\ \end{align}$$

Hint: Make another substitution $z=2t+3$. However ugly it turns out to be, you only need to calculate integrals of the form $x^{\alpha}$ for $\alpha$ integer.

You can first observe that $$ \sqrt{x^2-3x+2}=\frac{1}{2}\sqrt{4x^2-12x+8}=\frac{1}{2}\sqrt{(2x-3)^2-1} $$ so with $2x-3=t$, you get $$ \frac{1}{4}\int_3^5\sqrt{t^2-1}\,dt $$ Now use the Euler substitution $\sqrt{t^2-1}=t-u$, so $t^2-1=t^2-2tu+u^2$ and $t=\frac{u^2+1}{2u}$. Thus $$ 2t=u+\frac{1}{u},\qquad 2\,dt=\left(1-\frac{1}{u^2}\right)\,du=\frac{u^2-1}{u^2}\,du $$ and the integral becomes $$ \frac{1}{8}\int_{3-2\sqrt{2}}^{5-2\sqrt{6}}\frac{u^2-1}{u^2}\left(\frac{u^2+1}{2u}-u\right)\,du= -\frac{1}{16}\int_{3-2\sqrt{2}}^{5-2\sqrt{6}}\left(u-\frac{2}{u}+\frac{1}{u^2}\right)\,du $$ Not nice, but less ugly. Check the computations, please.