2.4 When the denominator has repeated quadratic factors

1.1 Proper fraction
1.2 Improper fraction

2.1 When the denominator has non-repeated linear factors
2.2 When the denominator has repeated linear factors
2.3 When the denominator has non-repeated quadratic factors
2.4 When the denominator has repeated quadratic factors

The quadratic polynomial of the form $\left( {p{x^2} + qx + r} \right)$ which is repeated $r$ times and cannot be factorized as linear factors in the denominator. Then the fraction is expressed as the sum of partial fraction as, $$ \Rightarrow \frac{{{A_1}x + {B_1}}}{{\left( {p{x^2} + qx + c} \right)}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {p{x^2} + qx + c} \right)}^2}}} + ... + \frac{{{A_r}x + {B_r}}}{{{{\left( {p{x^2} + qx + c} \right)}^r}}}$$

Question 7. Resolve $\frac{{x}}{{(x + 1){{({x^2} + 1)}^2}}}$ as partial fractions.

Solution: In the above fraction, we have ${{{({x^2} + 1)}^2}}$ as repeated quadratic factors which cannot be factorized into lienar factors.

So, the above fraction can be expressed as the sum of partial fraction as, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{x}}{{(x + 1){{({x^2} + 1)}^2}}} = \frac{A}{{(x + 1)}} + \frac{{Bx + C}}{{({x^2} + 1)}} + \frac{{Dx + E}}{{{{({x^2} + 1)}^2}}} \\ \Rightarrow \frac{{A{{({x^2} + 1)}^2} + (Bx + C)(x + 1)({x^2} + 1) + (Dx + E)(x + 1)}}{{(x + 1){{({x^2} + 1)}^2}}} = \frac{{x}}{{(x + 1){{({x^2} + 1)}^2}}} \\\end{aligned} \end{equation} $$ Equating the numerators we get, $$ \Rightarrow A{({x^2} + 1)^2} + (Bx + C)(x + 1)({x^2} + 1) + (Dx + E)(x + 1) = x$$

For $(x+1)=0$ or $x=-1$,$$\begin{equation} \begin{aligned} \Rightarrow A{(1 + 1)^2} + ( - B + C)( - 1 + 1)(1 + 1) + ( - D + E)( - 1 + 1) = - 1 \\ \Rightarrow 4A = - 1 \\ \Rightarrow A = - \frac{1}{4} \\\end{aligned} \end{equation} $$ For $({x^2} + 1)=0$ or ${x^2} = - 1$, $$\begin{equation} \begin{aligned} \Rightarrow A(0) + (Bx + C)(x + 1)(0) + (Dx + E)(x + 1) = x \\ \Rightarrow (Dx + E)(x + 1) = x \\ \Rightarrow D{x^2} + Dx + Ex + E = x \\ \Rightarrow - D + Dx + Ex + E = x\quad \quad \left( {As,\;{x^2} = - 1} \right) \\ \Rightarrow (D + E)x + (E - D) = x \\\end{aligned} \end{equation} $$

Equating the coeffiecient of RHS & LHS we get, $$\begin{equation} \begin{aligned} \Rightarrow D + E = 1\quad \quad \ \quad \quad E = D \\ \Rightarrow D = E = \frac{1}{2} \\\end{aligned} \end{equation} $$

For $x=0$, $$\begin{equation} \begin{aligned} \Rightarrow A + C + E = 0 \\ \Rightarrow - \frac{1}{4} + C + \frac{1}{2} = 0 \\ \Rightarrow C = - \frac{1}{4} \\\end{aligned} \end{equation} $$

For $x=1$, $$\begin{equation} \begin{aligned} \Rightarrow 4\left( { - \frac{1}{4}} \right) + \left( {B - \frac{1}{4}} \right)(4) + \left( {\frac{1}{2} + \frac{1}{2}} \right)(2) = 1 \\ \Rightarrow - 1 + 4B - 1 + 2 = 1 \\ \Rightarrow 4B = 1 \\ \Rightarrow B = \frac{1}{4} \\\end{aligned} \end{equation} $$

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