<s>[INST] Evaluate $\int \frac{x^2}{x^2 -6x + 10}\,dx$ Evaluate $$\int \frac{x^2}{x^2 -6x + 10} \, dx$$
I'd love to get a hint how to get rid of that nominator, or make it somehow simpler. 
Before posting this, I've looked into: Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$
And I've not understood how they simplied the nominator. I know that it has to match $2x-6$ somehow. but the way they put $(x-6)$ and multipled the integral and have suddenly in the nominator $x+1$ does not make sense to me.
 [/INST] Hint : $$\int\frac{x^2}{x^2-6x+10}dx=\int1+\frac{6x-10}{x^2-6x+10}dx$$
$\dfrac{d}{dx} \ln{(x^2-6x+10)}=\frac{2x-6}{x^2-6x+10}$ , therefore make 
$$\frac{6x-10}{x^2-6x+10}=\frac{3(2x-6)}{x^2-6x+10}+\frac{8}{x^2-6x+10}=3\frac{2x-6}{x^2-6x+10}+8\frac{1}{(x-3)^2+1}$$
$$I=x+3\ln(x^2-6x+10)+8\arctan{(x-3)}+constant$$
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