# Partial Fraction – Problems

### Partial Fraction – Problems

If f(x) and g(x) are two polynomials, then $$\frac{f\left( x \right)}{g\left( x \right)}$$ defines a rational algebraic function or a rational function of x.

Case I: when denominator is expressed is expressible as the product of non-repeating linear factors.

Let g(x) = (x – a₁) (x – a₂) … (x – an) then we assume that $$\frac{f\left( x \right)}{g\left( x \right)}=\frac{{{A}_{1}}}{x-{{a}_{1}}}+\frac{{{A}_{2}}}{x-{{a}_{2}}}+…+\frac{{{A}_{n}}}{x-{{a}_{n}}}$$.

1. Resolve $$\frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}$$ into Partial fractions.

Solution: Given that $$\frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}$$.

Let $$\frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}=\frac{A}{x+2}+\frac{B}{x-1}$$.

Then $$\frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}=\frac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}$$.

5x + 1 = A (x – 1) + B (x + 2) … (1)

Put x = 1 in (1)

5 (1) + 1 = A (0) + B (1 + 2)

3B = 6 ⇒ B = 2

Put x = – 2 in (1)

5 (-2) + 1 = A (- 2 – 1) + B (0)

– 9 = – 3A ⇒ A = 3

∴ $$\frac{5x+1}{\left( x+2 \right)\left( x-1 \right)}=\frac{3}{x+2}+\frac{2}{x-1}$$.

2. Resolve $$\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}$$ into Partial fractions.

Solution: Given that $$\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}$$.

Let $$\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}=\frac{A}{x+2}+\frac{B}{2x+1}$$.

$$\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{A\left( 2x+1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( 2x+1 \right)}$$.

2x + 3 = A (2x + 1) + B (x + 2) … (1)

Put x = -2 in (1)

2 (-2) + 3 = A [2 (-2) + 1] + B (0)

-1 = A (-3) ⇒ A = ⅓

Put x = – ½ in (1)

2 (- ½) + 3 = A (- 1 + 1) + B (- ½ + 2)

2 = B (3/2)

B = 4/3

∴ $$\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{1/3}{x+2}+\frac{4/3}{2x+1}$$.

∴ $$\frac{2x+3}{\left( x+2 \right)\left( 2x+1 \right)}=\frac{1}{3\left( x+2 \right)}+\frac{4}{3\left( 2x+1 \right)}$$.

Hence $$\frac{2x+3}{5\left( x+2 \right)\left( 2x+1 \right)}=\frac{1}{15\left( x+2 \right)}+\frac{4}{15\left( 2x+1 \right)}$$.

3. Resolve $$\frac{{{x}^{2}}+5x+7}{{{\left( x-3 \right)}^{3}}}$$ into Partial fractions.

Solution: Given that $$\frac{{{x}^{2}}+5x+7}{{{\left( x-3 \right)}^{3}}}$$.

Let x – 3 = y ⇒ x = y + 3

$$\frac{{{x}^{2}}+5x+7}{{{\left( x-3 \right)}^{3}}}=\frac{{{\left( y+3 \right)}^{2}}+5\left( y+3 \right)+7}{{{y}^{3}}}$$.

$$=\frac{{{y}^{2}}+6y+9+5y+15+7}{{{y}^{3}}}$$.

$$=\frac{{{y}^{2}}+11y+31}{{{y}^{3}}}=\frac{1}{y}+\frac{11}{{{y}^{2}}}+\frac{31}{{{y}^{3}}}$$.

∴ $$\frac{{{x}^{2}}+5x+7}{{{\left( x-3 \right)}^{3}}}=\frac{1}{x-3}+\frac{11}{{{\left( x-3 \right)}^{2}}}+\frac{31}{{{\left( x-3 \right)}^{3}}}$$.