I have the following integral :
$$\int \frac{x^2+2x+3}{x^2-5x+6}~dx$$
Although I tried to solve it by partial fractions, I could not come up with an appropriate answer. My question is more about the technique which I should use here in order to evalue the integral. I understand that the denominator could be expressed as $$(x-2)(x-3)$$ But then how to proceed with such a complicated numerator.
First, $$ x^2+2x+3=1\cdot (x^2-5x+6)+7x-3. $$ Thus $$ \frac{x^2+2x+3}{x^2-5x+6}=1+\frac{7x-3}{x^2-5x+6}. $$ You know $x^2-5x+6=(x-2)(x-3)$. It is known that there exists $A,B$ such that $$ \frac{7x-3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3} $$ and so $$ \frac{7x-3}{x^2-5x+6}=\frac{(A+B)x-(3A+2B)}{x^2-5x+6}. $$ Thus we get $A=-11$ and $B=18$. Therefore $$ \int\frac{x^2+2x+3}{x^2-5x+6}dx=x-11\ln|x-2|+18\ln|x-3|+C. $$