Question 326720
Using partial fractions we have
(7x + 1)/[(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2)


Multiply throughout by (x + 1)(x - 2) gives
7x + 1 = A(x - 2) + B(x + 1)


By putting x=2 we can eliminate A and get
7(2) + 1 = A(2 - 2) + B(2 + 1)
15 = 3B
B = 5


Similarly by putting x=-1 we can eliminate B and get
7(-1) + 1 = A(-1 - 2) + B(-1 + 1)
-6 = -3A
A = 2


Therefore (7x + 1)/[(x + 1)(x - 2)] = 2/(x + 1) + 5/(x - 2)


Check:

(7x + 1)/[(x + 1)(x - 2)] = [2(x - 2) + 5/(x + 1)]/[(x + 1)(x - 2)]

(2x - 4 + 5x + 5)/[ (x + 1)(x - 2)] = (7x + 1)/[(x + 1)(x - 2)]


So using this example should enable you to work out other fractions like this using this method.
It is a bit more difficult for repeated factors and factors of higher orders but for the moment we will stick to non-repeated linear factors.