Partial Fraction

Type 1: If $\frac{x-1}{(x+1{)}^2}$ = $\frac{A}{x+1}$ + $\frac{B}{(x+1{)}^2}$ , then value of A and B are

Step 1: 

set $(x+1{)}^2=0$ and solve for x. In this case x = -1. This trick works because the denominator of the partial fraction is the same as denominator of the original fraction.

Now plug in x = -1 into the numerator of the original fraction to find the value of B. In this case, B = -1-1 = -2

To find the value of A, plug in x=0 in the whole equation given and calculate the value of A. In this case, $\frac{0-1}{(0+1{)}^2}=\frac{A}{0+1}+\frac{-2}{(0+1{)}^2}$ or, -1 = A - 2 or, A = -1 + 2 = 1

Type 2: If $\frac{3x-7}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)}$ , then value of A and B are

Set x-2 = 0 and solve for x. In this case, x = 2. Now put x = 2 in the original fraction except for (x-2) to find the value of A. 

$$A=\frac{3\times 2-7}{2-3}=\frac{6-7}{-1}=\frac{-1}{-1}=1$$ Step 2:

Set x-3 = 0 and solve for x. In this case, x = 3. Now put x = 3 in the original fraction except for (x-3) to find the value of B.

$$B=\frac{3\times 3-7}{3-2}=\frac{9-7}{-1}=\frac{-2}{-1}=2$$