### A problem asked by Daniel Phulbani

Evaluate $\int \frac{x - 1}{\sqrt{x^2 - 2x + 4}} \;dx$

Substitute $u = x^2 - 2x + 4$ and $du = (2x-2)\;dx = 2(x-1) \;dx$ and get

$\int \frac{x - 1}{\sqrt{x^2 - 2x + 4}} \;dx$

$= \frac{1}{2} \int \frac{1}{\sqrt{x^2 - 2x + 4}} \cdot 2(x-1)\;dx$

$= \frac{1}{2} \int \frac{1}{\sqrt{u}} \;du$

$= \frac{1}{2} \int u^{\frac{-1}{2}} \;du$

$= \frac{1}{2} \frac{u^{\frac{1}{2}}} {\frac{1}{2}} + C$

$= \sqrt{u} + C = \sqrt{x^2-2x+4} + C$