Question

The partial fraction decomposition of \( \frac{x^{4}+24 x^{2}+28}{(x^{2}+1)^{3}} \) is

• A. \( \frac{1}{x^{2}+1}-\frac{22}{(x^{2}+1)^{2}}+\frac{5}{(x^{2}+1)^{3}} \)
• B. \( \frac{1}{x^{2}+1}+\frac{22}{(x^{2}+1)^{2}}+\frac{5}{(x^{2}+1)^{3}} \)
• C. \( \frac{1}{x^{2}+1}-\frac{22}{(x^{2}+1)^{2}}-\frac{5}{(x^{2}+1)^{3}} \)
• D. \( \frac{1}{x^{2}+1}+\frac{22}{(x^{2}+1)^{2}}-\frac{5}{(x^{2}+1)^{3}} \)

Hint:

For partial fractions with repeating powers of a single expression, substitution is always faster than the method of equating coefficients. It turns a complex algebra problem into a simple polynomial expansion!

Answer 1

The Correct Option is B

Concept: When a rational function contains a repeated quadratic factor like \( (x^2 + 1)^3 \) in the denominator, we can perform a simple substitution to easily find the partial fraction decomposition instead of solving a tedious system of linear equations. Let \( t = x^2 + 1 \), which implies \( x^2 = t - 1 \).

Step 1: Substituting \( x^2 = t - 1 \) into the numerator.
The numerator of the expression is \( x^4 + 24x^2 + 28 \). Since \( x^4 = (x^2)^2 = (t-1)^2 \), let us substitute: \[ \text{Numerator} = (t - 1)^2 + 24(t - 1) + 28 \] Expanding the squared term: \[ = (t^2 - 2t + 1) + 24t - 24 + 28 \] Combining like terms: \[ = t^2 + (-2t + 24t) + (1 - 24 + 28) = t^2 + 22t + 5 \]

Step 2: Dividing by the denominator \( t^3 \).
Now, place this simplified numerator back over the substituted denominator: \[ \frac{t^2 + 22t + 5}{t^3} = \frac{t^2}{t^3} + \frac{22t}{t^3} + \frac{5}{t^3} = \frac{1}{t} + \frac{22}{t^2} + \frac{5}{t^3} \]

Step 3: Substituting back \( t = x^2 + 1 \).
\[ = \frac{1}{x^2+1} + \frac{22}{(x^2+1)^2} + \frac{5}{(x^2+1)^3} \] This matches option (B) exactly.