Question

If \[ \frac{x+3}{(1-x)^2(1+x^2)} = \frac{A}{(1-x)} + \frac{B}{(1-x)^2} + \frac{Cx+D}{2(1+x^2)}, \] then \(B^2+C^2+D^2=\)

• A. \(\dfrac{3}{4}\)
• B. \(\dfrac{9}{2}\)
• C. \(14\)
• D. \(4\)

Hint:

For repeated linear factors in partial fractions, \[ \boxed{\text{Substitute the repeated root first to obtain the highest power coefficient.}} \] The remaining constants can then be found using differentiation or coefficient comparison.

Answer 1

The Correct Option is C

Step 1: Find \(B\). Multiplying throughout by \((1-x)^2\) and putting \[ x=1, \] we get \[ B=\frac{1+3}{1+1^2}=2. \]

Step 2: Find \(A\). Differentiate \[ \frac{x+3}{1+x^2} =A(1-x)+B +\frac{(Cx+D)(1-x)^2}{2(1+x^2)}, \] and substitute \[ x=1. \] This gives \[ A=\frac12. \]

Step 3: Find \(C\) and \(D\). Using \[ \frac{x+3}{1+x^2} =\frac12(1-x)+2+\frac{Cx+D}{2(1+x^2)}(1-x)^2, \] and comparing coefficients, we obtain \[ C=1,\qquad D=3. \] Hence, \[ B^2+C^2+D^2 =2^2+1^2+3^2 =4+1+9 =14. \] Therefore, \[ \boxed{14}. \] Thus, \[ \boxed{(C)} \] is the correct answer.