The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;
#(5x-1)/((x-2)(x+1)) = A/(x-2) + B/(x+1)#
Where we need to solve for #A# and #B#. We can cross multiply to combine the terms on the right hand side over a common denominator. We get;
#(5x-1)/((x-2)(x+1)) = (A(x+1) + B(x-2))/((x-2)(x+1))#
We can now cancel the denominator on each side, leaving;
#5x-1 = A(x+1) + B(x-2)#
Now we can solve for #A# and #B#. We can make one of the terms cancel out by choosing the right value for #x#. Lets try #x=~1#.
#5(~1)-1 = A(~1+1) + B(~1-2)#
The #A# term goes away since it is multiplied by zero, leaving;
#~6 = ~3B#
Solving for #B#;
#B=2#
We can substitute #B# and solve for #A#, but it would be easier to do the same trick that we used to solve for #B#. Let #x=2#.
#5(2) -1 = A(2+1) + B(2-2)#
This time, the #B# term goes away;
#9 = 3A#
#A = 3#
Now that we have our values for #A# and #B# we can plug into our first function and get;
#(5x-1)/((x-2)(x+1)) = 3/(x-2) + 2/(x+1)#
