I've been stuck on this problem for 2 hours and I have no idea where I'm going wrong. Here is the original equation:
$$\int_{0}^{1}\frac{x-4}{x^2-5x+6}dx$$
I first factored and rewrote the fraction this way:
$$\frac{A(x-1)+B(x-6)}{(x-6)(x+1)}$$
After expanding and grouping I got this:
$$\frac{(A+B)x+(A-6B)}{(x-6)(x+1)}$$
I then solved the system of equations by means of algebraic substitution and got $A=\frac{2}{7}$ and $B=\frac{5}{7}$.
I rewrote my integral using my new $A$ and $B$ values:
$$\int_{0}^{1}\frac{2}{7x-42}+\frac{5}{7x+7}dx$$
I tried integrating the above by factoring out a $\frac{1}{7}$ from both terms and I tried integrating it in unfactored form. Neither answer was correct.
In factored form I got the following:
$$\frac{1}{7}(2ln|x-6|)+\frac{1}{7}(5ln|x+1|)|_{0}^{1}$$
In unfactored form I got the following:
$$14ln|7x-42|+35ln|7x+7||_{0}^{1}$$
Is everything up to this point correct and my error with the evaluation of the final step or did I make a mistake earlier on?