<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] You can use long division. Another way is as follows:
\begin{align*}
\frac{x^2+1}{x^2-5x+6}&=\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}\\
&=\frac{x^2-5x+6}{x^2-5x+6}+\frac{5x-5}{x^2-5x+6}\\
&=1+\frac{5x-5}{x^2-5x+6}
\end{align*}
As next step notice that $$x^2-5x+6=(x-2)(x-3)$$
and expand in partial fractions:
\begin{align*}
\frac{5x-5}{x^2-5x+6}&=\frac{A}{x-2}+\frac{B}{x-3}&&\text{Where }A\text{ and }B\text{ are constants}
\end{align*}
</s>