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.  

Given $\displaystyle \int\frac{x^2+1}{x^2-5x+6}dx = \int\frac{(x^2-5x+6)+(5x-5)}{x^2-5x+6}dx = \int 1dx+5\int\frac{(x-1)}{(x-2)(x-3)}dx$
Now Using Partial fraction Method::
$\displaystyle \frac{x-1}{(x-2)\cdot (x-3)} = \frac{A}{(x-2)}+\frac{B}{(x-3)}$
So we get $x-1 = A(x-3)+B(x-2) = (A+B)x+(-3A-2B)$
After Camparing the coefficient, we get $\displaystyle A+B=1$ and $3A+2B = 1$
So we get $\displaystyle A = -1$ and $\displaystyle B= 2$
So we get $\displaystyle \frac{x^2+1}{x^2-5x+6}dx = x+5\int \left[-\frac{1}{x-2}+\frac{2}{x-3}\right]dx$
