Question 12135
{{{ (5x + 7)/ (x^2 + 2x -3) }}}
{{{ (5x + 7)/ ((x + 3)(x -1)) = A/(x+3) + B/(x-1)}}}


Multiply both sides of this equation by the LCD, which is (x+3)(x-1).

The result will be:
5x + 7 = A(x-1) + B(x+3)


You can substitute any two values of x into this equation, and obtain two equations and two unknowns.  However, the easiest way to do it is to let x=1 and let x= -3, since these particular values of x will "zero out" half of the problem and give immediate results:

If x= 1
5x + 7 = A(x-1) + B(x+3)
5 + 7 = A(0) + B(1+3)
12=4B, so B= 3


If x = -3,
5x + 7 = A(x-1) + B(x+3)
-15+7 = A(-3-1) + B(0)
-8 = -4A, so A = 2


Therefore the partial fractions would be:
{{{ (5x + 7)/ ((x + 3)(x -1)) = A/(x+3) + B/(x-1)}}}
{{{ (5x + 7)/ ((x + 3)(x -1)) = 2/(x+3) + 3/(x-1)}}}


R^2 at SCC