Find $\int\frac{x-1}{x^2-5x+6}dx$. Why my solution is different from book? I'm learning single variable calculus right now. Right now trying to understand integration with partial fraction. I'm confused in a problem from sometime. I think I'm doing right but answer in my book is something else. 
Please have a look at the images. 



The Solution given in my Book  
I know there's is a difference in finding value of A and B. But in previous exercise I was applying the same method and was getting correct answer. Please help. thankyou in advance.
HINT: we get $$\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}$$
can you show this?
now we calculate the zeros of $x^2-5x+6$, these are:$$x_{1,2}=\frac{5}{2}\pm\sqrt{\frac{25}{4}-\frac{24}{4}}$$ thus we get $$x_1=3$$ or $$x_2=2$$
and we get $$x^2-5x+6=(x-2)(x-3)$$ and we can make the ansatz:
$$\frac{5x-5}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}$$
multiplying by the denominators we obtain:
$$\frac{5x-5}{x^2-5x+6}=\frac{A(x-3]+B(x-2)}{x^2-5x+6}$$
and we get $$5x-5=x(A+B)-3A-2B$$
from here you will get $$5=A+B$$ and $$-5=-3A-2B$$ you must solve this system
If your way is right, then the answer doesn't matter, because it is the representation of  the value, if you put the value and plot it on a garph, you will see you will get the same answer.