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.

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.
