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.

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.
