Find $\int\frac{x-1}{x^2-5x+6}dx$. Why my solution is different from book?

Question

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.

Answer 1

I think I get the answer of my question from where I was wrong.

Answer 2

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

Answer 3

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.

Answer 4

For the first part, I see what you're doing. $$\frac{x-1}{x^2-5x+6}=\frac{A\frac{d}{dx}(x^2-5x+6)+B}{x^2-5x+6}=\frac{A(2x-5)+B}{x^2-5x+6}$$ Which sets up the system of equations $$2A=1$$ $$-5A+B=-1$$ Another approach is just to perform algebraic manipulations. It may make the problem seem less procedural.

$$\frac{x-1}{x^2-5x+6}=\frac{\frac{1}{2}(2x-2)}{x^2-5x+6}=\frac{\frac{1}{2}(2x-5+5-2)}{x^2-5x+6}=\frac{2x-5}{2(x^2-5x+6)}+\frac{3}{2(x^2-5x+6)}$$ In any event, you're still left with the same problem. While the first fraction can be integrated with the substitution $u=x^2-5x+6$, the second one still has a quadratic denominator that needs to be handled some how. Obvious methods include integration by parts, or inverse trigonometric substitution (circular will do, but hyperbolic is cleaner for the $x^2-1$ form).