Question

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

• A. 0
• B. 1
• C. 2
• D. -1

Hint:

When dealing with partial fractions involving repeated linear factors like $(x-a)^2$, remember to include terms for each power: $\frac{A}{x-a} + \frac{B}{(x-a)^2}$. The "cover-up" method works for the highest power (finding B), but other methods like substituting values or equating coefficients are needed for the other constants.

Answer 1

The Correct Option is D

Step 1: Partial fraction decomposition.
Multiply both sides by $(x-1)^2(x^2+1)$: \[ 3x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2. \]
Step 2: Find $B$ using $x=1$.
\[ 3(1)+1 = 0 + 2B + 0 \implies 4 = 2B \implies B=2. \]
Step 3: Expand and collect coefficients.
\[ 3x+1 = (A+C)x^3 + (-A+B-2C+D)x^2 + (A+C-2D)x + (-A+B+D) \]
Step 4: Solve for $A$, $C$, $D$.
\[ A+C=0 \implies C=-A, -A+B+D=1 \implies -A+2+D=1 \implies D=A-1 \] \[ -A+B-2C+D=0 \implies -A+2-2(-A)+D=0 \implies 3A+D+2=0 \implies D=-3/2, A=-1/2 \] \[ C=-A=1/2 \]
Step 5: Compute $2(A-C+B+D)$.
\[ 2(A-C+B+D) = 2\Big(-\frac12 - \frac12 + 2 - \frac{3}{2}\Big) = -1 \]