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.

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

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




Partial Fractions

Chapter 5: Decomposition (Type-4)


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