$$\frac{(7x+2)(x+1)}{(x^2-1)(x+2)} = \frac A{x−1}+\frac B{x+2}$$
I never did such a task before. My first idea would be to shorten the left side by (x+1), which would result in.
$$\frac{7x+2}{(x^2-1)+1} = \frac A{x−1}+\frac B{x+2}$$ And that is utterly wrong. I asked in other forums and they said me that I have to get this term:
$$\frac{7x+2}{(x-1)(x+2)} = \frac {A(x+2)+B(x-1)}{(x-1)(x+2)}$$ With the solutions being $$(A+B)*x = 7x~and~ 2A - B = 2$$
I did not understand anything, how to get to these results in the first instance and how to calculate A and B. This is my one of my first "tasks" I have to do for the first semester, and I have no idea how to do it. Need help! =(
After you get
$$7x+2=A(x+2)+B(x-1)$$
rewwrite it as
$$7x+2=(A+B)x+(2A-B)$$
Hence by comparing the coefficient, you just have to solve for $A+B=7$ and $2=2A-B$.
$$2=2A-(7-A)=3A-7$$
$$3A=2+7=9$$
Hopefully you can solve for $A$ and $B$ from here.
Alternatively, after getting
$$7x+2=A(x+2)+B(x-1)$$
Let $x=1$, and we get $9=3A$ immediately.
Let $x=-2$, we get $-14+2=-3B$ immediately.