**When the factors in the denominator are linear and repetitive:** Let the linear factors in the denominator are \((x - a)\), and \((x - b)^2\).

Here \((x - b)\) is the repetitive term which came two times, hence write the fractions like the given below and should be equal to the main fraction then solve the equation. $$ \frac{A}{(x - a)} + \frac{B}{(x - b)} + \frac{C}{(x - b)^2}.... $$ Here A, B, and C are the unknown constants.

**Note(1):** If there is a quadratic equation in the denominator, in that case, we have to factorize the quadratic equation into linear factors first. If it is not possible to factorize the quadratic equation then use the type III or Type IV method to solve the question.

**Note(2):** We can also assume the repetitive term equal to \(y\) and then solve the equation easily. We can better understand the concept by the second example given below.

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

**Solution: Method(1):** The given equation can also be written as

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{A}{(x - 2)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2}....(1)\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{A}{(x - 2)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2}....(1) $$

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

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)}{(x - 2) (x - 1)^2}\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)}{(x - 2) (x - 1)^2} $$

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

\(2x^2 - x + 1 =\) \(A (x - 1)^2 + B (x - 2) \ (x - 1) +\) \(C (x - 2)....(2)\)

$$ 2x^2 - x + 1 = A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)....(2) $$

Now to calculate the value of A, we can take the denominator of A from the equation (1) and put it equal to zero (0). $$ x - 2 = 0 $$ $$ x = 2 $$ Now put this value of \(x\) in equation (2).

\(2 \ (2)^2 - 2 + 1 =\) \(A \ (2 - 1)^2 + B \ (2 - 2) \ (2 - 1) +\) \(C \ (2 - 2)\)

$$ 2 \ (2)^2 - 2 + 1 = A \ (2 - 1)^2 + B \ (2 - 2) \ (2 - 1) + C \ (2 - 2) $$

$$ 8 - 2 + 1 = A + 0 + 0 $$ $$ A = 7 $$ Similarly to calculate the value of C, we can take the denominator of C from equation (1) and put it equal to zero (0). $$ (x - 1)^2 = 0 $$ $$ x - 1 = 0 $$ $$ x = 1 $$ again put this value of \(x\) in equation (2).

\(2 \ (1)^2 - 1 + 1 =\) \(A (1 - 1)^2 + B (1 - 2) \ (1 - 1) + C (1 - 2)\)

$$ 2 \ (1)^2 - 1 + 1 = A (1 - 1)^2 + B (1 - 2) \ (1 - 1) + C (1 - 2) $$

$$ 2 = 0 + 0 + C \ (-1) $$ $$ C = -2 $$

Now to calculate the value of B, we can put \(x = 0\) in equation (2).

\(2 \ (0)^2 - 0 + 1 =\) \(A (0 - 1)^2 + B (0 - 2) \ (0 - 1) + C (0 - 2)\)

$$ 2 \ (0)^2 - 0 + 1 = A (0 - 1)^2 + B (0 - 2) \ (0 - 1) + C (0 - 2) $$

$$ 1 = A \ (-1)^2 + B \ (-2) \ (-1) + C \ (-2) $$ $$ 1 = A + 2B - 2C $$ Now putting the values of A and C in the equation. $$ 1 = 7 + 2B - 2 \ (-2) $$ $$ 1 = 7 + 2B + 4 $$ $$ 1 = 11 + 2B $$ $$ 2B = 1 - 11 $$ $$ 2B = -10 $$ $$ B = -5 $$

Finally, put the values of A, B, and C in equation (1).

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{7}{(x - 2)} + \frac{-5}{(x - 1)} + \frac{-2}{(x - 1)^2}\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{7}{(x - 2)} + \frac{-5}{(x - 1)} + \frac{-2}{(x - 1)^2} $$

Hence this is the final decomposition of the partial fraction equation.

**Solution: Method(2):** The given equation can also be written as

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{A}{(x - 2)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2}....(1)\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{A}{(x - 2)} + \frac{B}{(x - 1)} + \frac{C}{(x - 1)^2}....(1) $$

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

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)}{(x - 2) (x - 1)^2}\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)}{(x - 2) (x - 1)^2} $$

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

\(2x^2 - x + 1 =\) \(A (x - 1)^2 + B (x - 2) \ (x - 1) +\) \(C (x - 2)....(2)\)

$$ 2x^2 - x + 1 = A (x - 1)^2 + B (x - 2) \ (x - 1) + C (x - 2)....(2) $$

Write down the terms of \(x^2\), \(x\), and constants separately as written below.

\(2x^2 - x + 1 =\) \(A (x^2 + 1 - 2x) + B (x^2 - x - 2x + 2) +\) \(C (x - 2)\)

$$ 2x^2 - x + 1 = A (x^2 + 1 - 2x) + B (x^2 - x - 2x + 2) + C (x - 2) $$

\(2x^2 - x + 1 =\) \(A (x^2 + 1 - 2x) + B (x^2 - 3x + 2) +\) \(C (x - 2)\)

$$ 2x^2 - x + 1 = A (x^2 + 1 - 2x) + B (x^2 - 3x + 2) + C (x - 2) $$

\(2x^2 - x + 1 =\) \(x^2 \ (A + B) + x \ (-2A - 3B + C)\) \(+ (A + 2B - 2C)\)

$$ 2x^2 - x + 1 = x^2 \ (A + B) + x \ (-2A - 3B + C) + (A + 2B - 2C) $$

Now compare the values of \(x^2\), \(x\), and constants, of RHS and LHS. $$ A + B = 2....(3) $$ $$ -2A - 3B + C = -1....(4) $$ $$ A + 2B - 2C = 1....(5) $$ After multiplying by 2 with equation (4) add it with equation (5).

\(-4A - 6B + 2C +\) \(A + 2B - 2C = -2 + 1\)

$$ -4A - 6B + 2C + A + 2B - 2C = -2 + 1 $$

$$ -3A -4B = -1....(6) $$ Now multiply by 4 with equation (3) and add it with equation (6). $$ 4A + 4B - 3A - 4B = 8 - 1 $$ $$ A = 7 $$ Now put the value of A in equation (3). $$ 7 + B = 2 $$ $$ B = -5 $$ Now put the value of A and B in equation (4). $$ -2 \times 7 - 3 \ (-5) + C = -1 $$ $$ -14 + 15 + C = -1 $$ $$ 1 + C = -1 $$ $$ C = -2 $$ Finally, put the values of A, B, and C in equation (1).

\(\frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} =\) \(\frac{7}{(x - 2)} + \frac{-5}{(x - 1)} + \frac{-2}{(x - 1)^2}\)

$$ \frac{2x^2 - x + 1}{(x - 2) \ (x - 1)^2} = \frac{7}{(x - 2)} + \frac{-5}{(x - 1)} + \frac{-2}{(x - 1)^2} $$

Hence this is the final decomposition of the partial fraction equation.

**Example(2):** \(\frac{x^2}{(x - 1)^3}\)

**Solution:** Let \(x - 1 = y\) then \(x = y + 1\) then $$ \frac{x^2}{(x - 1)^3} = \frac{(y + 1)^2}{y^3} $$ $$ = \frac{y^2 + 2y + 1}{y^3} $$ It can also be written as. $$ = \frac{1}{y} + \frac{2}{y^2} + \frac{1}{y^3} $$ After putting the value of \(y\) we get.

\(\frac{x^2}{(x - 1)^3} =\) \(\frac{1}{(x - 1)} + \frac{2}{(x - 1)^2} + \frac{1}{(x - 1)^3}\)

$$ \frac{x^2}{(x - 1)^3} = \frac{1}{(x - 1)} + \frac{2}{(x - 1)^2} + \frac{1}{(x - 1)^3} $$

Hence this is the final decomposition of the partial fraction equation.

Lec 1: Introduction
Lec 2: Partial Fraction Decomposition (Type-1)
Lec 3: Partial Fraction Decomposition (Type-2)
Lec 4: Partial Fraction Decomposition (Type-3)
Lec 5: Partial Fraction Decomposition (Type-4)