<s>[INST] How would you solve $\int \frac{x}{x^2 - 4x + 5} dx$ What is the tip for integrating that integral? I completed the square on the bottom to make it $$\frac{x}{(x-2)^2  + 1}$$
but it doesn't seem helpful. Any tips? 
Thanks.
 [/INST] You can continue by letting $u=x-2$, $x=u+2$, $dx=du$ to get
$\displaystyle\int\frac{u+2}{u^2+1} du=\frac{1}{2}\int\frac{2u}{u^2+1} du+2\int\frac{1}{u^2+1} du=\frac{1}{2}\ln(u^2+1)+2\arctan u + C$
$\;\;=\frac{1}{2}\ln(x^2-4x+5)+2\arctan(x-2)+C$
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