Question

If \[ \frac{x}{(x-1)(x^2+1)^2} = \frac{1}{4} \left[\frac{1}{x-1} - \frac{x+1}{x^2+1}\right] + y, \] then \(y =\)

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

Hint:

When decomposing rational expressions, carefully align denominators and simplify stepwise to isolate unknown parts.

Answer 1

The Correct Option is A

Step 1: Set up the full partial fraction form.
Write \(\frac{x}{(x-1)(x^2+1)^2}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}\), then clear denominators:
\[ x = A(x^2+1)^2+(Bx+C)(x-1)(x^2+1)+(Dx+E)(x-1) \]

Step 2: Find A by plugging in x=1.
\(1=A(2)^2=4A \Rightarrow A=\frac{1}{4}\)

Step 3: Match coefficients of the remaining powers of x.
Expanding everything and comparing coefficients of \(x^4,x^3,x^2,x^0\) gives \(B=-\frac14\), \(C=-\frac14\), \(D=-\frac12\), \(E=\frac12\), and the leftover \(x^1\) coefficient checks out to 1, confirming these values.

Step 4: Read off y.
\[ \frac{x}{(x-1)(x^2+1)^2}=\frac14\left[\frac{1}{x-1}-\frac{x+1}{x^2+1}\right]+\frac{1-x}{2(x^2+1)^2} \]
Comparing with the given equation, y is exactly the leftover piece:
\[ \boxed{y=\frac{1-x}{2(x^2+1)^2}} \]