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

My try:
$$\sqrt {x^2-3x+2}=x+t$$
$$x=\frac{2-t^2}{2t+3}$$
$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$
$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$
$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$ However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.Can you help me?P.S. I must use Euler's substitution because that's the command.
 [/INST] Using the third substitution of Euler
$$\sqrt{{{x}^{2}}-3x+2}=\sqrt{\left( x-1 \right)\left( x-2 \right)}=\left( x-1 \right)t$$
$$x=\frac{2-{{t}^{2}}}{1-{{t}^{2}}}\Rightarrow dx=\frac{2t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}dt$$
$$\begin{align}
  & =-\int_{1/\sqrt{2}}^{\sqrt{2/3}}{\frac{2{{t}^{2}}}{{{\left( {{t}^{2}}-1 \right)}^{3}}}dt} \\ 
 & =\left[ \frac{\left( t+{{t}^{3}} \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}+\ln \left( \frac{1-t}{1+t} \right) \right]_{1/\sqrt{2}}^{\sqrt{2/3}} \\ 
\end{align}$$
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