<s>[INST] Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper rational fraction we have to divide
$$\frac{x^2 + 1}{x^2-5x+6}$$
I'm trying from sometime but couldn't find the right solution. 
 Please help! Thank you in advance.  
 [/INST] Notice, $$\frac{x^2+1}{x^2-5x+6}=\frac{(x^2-5x+6)+5(x-1)}{x^2-5x+6}$$
$$=1+5\frac{x-1}{x^2-5x+6}$$
$$\implies \frac{x-1}{x^2-5x+6}
=\frac{x-1}{(x-2)(x-3)}$$   $$=\frac{A}{x-2}+\frac{B}{x-3}$$
By comparing the corresponding coefficients, we get $A=-1$, $B=2$, hence 
$$\frac{x-1}{x^2-5x+6}=-\frac{1}{x-2}+\frac{2}{x-3}$$ Hence, we have  $$\frac{x^2+1}{x^2-5x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$$
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