ExamBro
ExamBro
AP EAMCET · Maths · Basic of Mathematics

\(\frac{x^4}{\left(x^2+1\right)\left(x^2+3\right)}=\)

  1. A \(\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+3}\) for some \(A, B, C, D \in R \backslash\{0\}\)
  2. B \(\frac{A x+B}{x^2+1}+\frac{C x}{x^2+1}\) for some \(A, B, C \in R \backslash\{0\}\)
  3. C \(\frac{A x}{x^2+1}+\frac{B x}{x^2+3}\) for some \(A, B \in R \backslash\{0\}\)
  4. D \(1+\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+3}\) for some \(A, B, C, D \in R\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1+\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+3}\) for some \(A, B, C, D \in R\)

Step-by-step Solution

Detailed explanation

\begin{aligned} \frac{x^4}{\left(x^2+1\right)\left(x^2+3\right)} & =\frac{\left(x^2+1\right)\left(x^2+3\right)-3 x^2-x^2-3}{\left(x^2+1\right)\left(x^2+3\right)} \\ & =1-\frac{4 x^2+3}{\left(x^2+1\right)\left(x^2+3\right)} \\ & =1+\frac{A x+B}{x^2+1}+\frac{C…