$$ \frac{x}{(x+1)(x,-1)^{3}}= $$ Restove pertial fraction

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To decompose the rational function into partial fractions, let's start with the given expression:

$$
\frac{x}{(x+1)(x-1)^3}
$$

**Step 1: Set Up the Partial Fraction Decomposition**

Given the denominator $$(x+1)(x-1)^3$$, the partial fractions will take the form:

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

**Step 2: Clear the Denominator**

Multiply both sides by $$(x+1)(x-1)^3$$ to eliminate the denominators:

$$
x = A(x-1)^3 + B(x+1)(x-1)^2 + C(x+1)(x-1) + D(x+1)
$$

**Step 3: Expand and Simplify**

Expand each term on the right-hand side:

$$
\begin{align*}
A(x-1)^3 &= A(x^3 - 3x^2 + 3x - 1) \
B(x+1)(x-1)^2 &= B(x^3 - x^2 - x + 1) \
C(x+1)(x-1) &= C(x^2 - 1) \
D(x+1) &= D(x + 1)
\end{align*}
$$

Combine all terms:

$$
x = (A + B)x^3 + (-3A - B + C)x^2 + (3A - B + D)x + (-A + B - C + D)
$$

**Step 4: Equate Coefficients**

Set the coefficients of corresponding powers of $$x$$ equal on both sides:

1. $$x^3$$: $$A + B = 0$$
2. $$x^2$$: $$-3A - B + C = 0$$
3. $$x$$: $$3A - B + D = 1$$
4. Constant term: $$-A + B - C + D = 0$$

**Step 5: Solve the System of Equations**

From equation (1):

$$
A + B = 0 \implies B = -A
$$

Substitute $$B = -A$$ into equation (3):

$$
3A - (-A) + D = 1 \implies 4A + D = 1
$$

From equation (4):

$$
-A + (-A) - C + D = 0 \implies -2A - C + D = 0
$$

And from equation (2):

$$
-3A - (-A) + C = 0 \implies -2A + C = 0 \implies C = 2A
$$

Substitute $$C = 2A$$ into the modified equation (4):

$$
-2A - 2A + D = 0 \implies -4A + D = 0 \implies D = 4A
$$

Now, substitute $$D = 4A$$ into equation $$4A + D = 1$$:

$$
4A + 4A = 1 \implies 8A = 1 \implies A = \frac{1}{8}
$$

Using $$A = \frac{1}{8}$$:

$$
B = -\frac{1}{8}, \quad C = 2 \times \frac{1}{8} = \frac{1}{4}, \quad D = 4 \times \frac{1}{8} = \frac{1}{2}
$$

**Step 6: Write the Final Partial Fraction Decomposition**

Substitute $$A$$, $$B$$, $$C$$, and $$D$$ back into the partial fractions:

$$
\frac{x}{(x+1)(x-1)^3} = \frac{1}{8(x+1)} - \frac{1}{8(x-1)} + \frac{1}{4(x-1)^2} + \frac{1}{2(x-1)^3}
$$

**Final Answer:**

$$
\frac{x}{(x+1)(x-1)^3} = \frac{1}{8(x + 1)} - \frac{1}{8(x - 1)} + \frac{1}{4(x - 1)^2} + \frac{1}{2(x - 1)^3}
$$
$$ \frac{x}{(x+1)(x-1)^3} = \frac{1}{8(x + 1)} - \frac{1}{8(x - 1)} + \frac{1}{4(x - 1)^2} + \frac{1}{2(x - 1)^3} $$

\( \frac{x}{(x+1)(x,-1)^{3}}= \) Restove pertial fraction

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