Question: Integral : $$\displaystyle \int \sqrt{x^2 - 5x + 4}~ dx$$
Found the solution, can't find step by step, the substitution $x-2=u$ was my first try, after that I am just stuck. :)

I am here: $\displaystyle \int \sqrt{u^2 -1} du$ and it seems so easy :)

Answer: 
Once you have $\int \sqrt{u^2-1}\, \mathrm{d}u$, try setting $u=\cosh(t)$ (or alternatively, $u=\sec(\theta)$).
First complete the square to obtain $\sqrt{(x-5/2)^2-9/4}=\frac12\sqrt{(2x-5)^2-9}.$

Then let $u=2x-5$ and therefore $du=2dx.$ So now your integral is:

$$\int\sqrt{u^2-9}\;du$$

Now envision a right triangle with acute angle $\theta$ whose opposite side is $\sqrt{u^2-9}.$ That makes the adjacent side $3$ and the hypotenuse $u,$ so that you can use $\sec\theta=u/3.$ Now you have $du=3\sec\theta\tan\theta,$ so that your integral becomes:

$$9\int\tan\theta \sec^2\theta\;d\theta.$$

From here, you will have to use the identity $1+\tan^2\theta=\sec^2\theta$ and integration by parts to get the final answer.

Good luck!