Question

The expression \[ \frac{x^4}{(x^2+1)(x^2+3)} \] is equal to

• A. \(\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}\) for some \(A,B,C,D\in \mathbb{R}\setminus\{0\}\)
• B. \(\dfrac{Ax+B}{x^2+1}+\dfrac{Cx}{x^2+1}\) for some \(A,B,C\in \mathbb{R}\setminus\{0\}\)
• C. \(\dfrac{Ax}{x^2+1}+\dfrac{Bx}{x^2+3}\) for some \(A,B\in \mathbb{R}\setminus\{0\}\)
• D. \(1+\dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{x^2+3}\) for some \(A,B,C,D\in \mathbb{R}\)

Hint:

If the degree of the numerator is equal to or greater than the degree of the denominator, first divide the polynomials. Then apply partial fractions to the proper rational part.

Answer 1

The Correct Option is D

Step 1: Compare degrees.
The denominator \((x^2+1)(x^2+3) = x^4+4x^2+3\) has degree 4, same as numerator \(x^4\). So the rational function is improper; perform polynomial division.

Step 2: Perform division.
\[\frac{x^4}{x^4+4x^2+3} = 1 - \frac{4x^2+3}{(x^2+1)(x^2+3)},\] and the remainder \(\tfrac{4x^2+3}{(x^2+1)(x^2+3)}\) decomposes as \(\tfrac{Ax+B}{x^2+1}+\tfrac{Cx+D}{x^2+3}\) for some constants \(A,B,C,D \in \mathbb{R}\). So the form is \(1 + \tfrac{Ax+B}{x^2+1}+\tfrac{Cx+D}{x^2+3}\).
\[\boxed{1 + \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3}}\]