# Partial Fractions Decomposition (Type-4):

When the quadratic factor in the denominator is repetitive: Let the quadratic factor in the denominator is $$(ax^2 + bx + c)^2$$, which is repetitive, hence we can write two fractions $$\frac{Ax + B}{(ax^2 + bx + c)}$$ and $$\frac{Cx + D}{(ax^2 + bx + c)^2}$$ equal to the main fraction. we can better understand by the given example.

Example: $$\frac{1}{(1 + x) \ (1 + x^2)^2}$$

Solution: According to the method the given equation can also be written as.

$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)}$$ $$+ \frac{Dx + E}{(1 + x^2)^2}....(1)$$

$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)} + \frac{Dx + E}{(1 + x^2)^2}....(1)$$

Now taking the LCM of RHS (Right Hand Side) terms.

$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)}{(1 + x) \ (1 + x^2)^2}$$

$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)}{(1 + x) \ (1 + x^2)^2}$$

The terms in the denominators on both sides are the same so they will be canceled.

$$1 = A (1 + x^2)^2 +$$ $$(1 + x) (1 + x^2) (Bx + C) +$$ $$(1 + x) (Dx + E)$$

$$1 = A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)$$

Write down the terms of $$x^4$$, $$x^3$$, $$x^2$$, $$x$$, and constants separately as written below.

$$1 = x^4 \ (A + B) + x^3 \ (B + C) +$$ $$x^2 \ (2A + B + C + D) +$$ $$x \ (B + C + D + E)$$ $$+ (A + C + E)$$

$$1 = x^4 \ (A + B) + x^3 \ (B + C) + x^2 \ (2A + B + C + D) + x \ (B + C + D + E) + (A + C + E)$$

After comparing the values of $$x^4$$, $$x^3$$, $$x^2$$, $$x$$, and constants of both sides we get. $$A + B = 0....(2)$$ $$B + C = 0....(3)$$ $$2A + B + C + D = 0....(4)$$ $$B + C + D + E = 0....(5)$$ $$A + C + E = 1....(6)$$ by subtracting the equation (5) from the equation (6). $$A - B - D = 1....(7)$$ Now by adding equation (4) and (7). $$3A + C = 1....(8)$$ Now by subtracting equation (3) from equation (2). $$A - C = 0....(9)$$ After solving the equations we get. $$A = \frac{1}{4}$$ $$B = \frac{-1}{4}$$ $$C = \frac{1}{4}$$ $$D = \frac{-1}{2}$$ $$E = \frac{1}{2}$$ After putting the values of A, B, C, D, and E in equation (1), we get the final equation of decomposition.

$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{1}{4 \ (1 + x)} +$$ $$\frac{(1 - x)}{4 \ (1 + x^2)} +$$ $$\frac{(1 - x)}{2 \ (1 + x^2)^2}$$

$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{1}{4 \ (1 + x)} + \frac{(1 - x)}{4 \ (1 + x^2)} + \frac{(1 - x)}{2 \ (1 + x^2)^2}$$