Find partial fractions of : (x+7) / ((x-3)(x+1)^2)
First, separate the algebraic fraction:(x+7) / ((x-3)(x+1)2 = A/(x+3) + B/(x+1) + C/(x+1)2 Multiply both sides by (x-3)(x+1)2: (x+7) = A(x+1)2+ B(x+1)(x+3) + C(x+3) Set x=-1 in order to eliminate A and B terms (-1+7)= 0 + 0 + C(-1+3) = 6 = 2C C=3 Set x=-3 in order to eliminate B and C terms: (-3+7)= A(-3+1)2 + 0 + 0 = 4 =-2A A=-2 Set x equal to a number that is not -1 or -3 and substitute A and C in order to find B I will set x=0 0+7 = -2(0+1)2 + B(0+1)(0+3) + 3(0+3) = 7 = -2 + 3B + 90 = 3B B = 0 Now substitute the values of A, B and C into the original seperated fraction (as B=0 that term will not appear in the answer): -2/(x+3) + 3/(x+1)2
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