The given equation in the image is:

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

We need to determine the constants $A$, $B$, $C$, and $D$.

Step-by-Step Solution:

1. Rewrite the equation: The left-hand side is a single fraction with the denominator $(x - 1)^2 (x^2 + 1)$, and the right-hand side has the same denominator expressed as a sum of partial fractions.

2. Combine the right-hand side under a common denominator: The right-hand side becomes: $\frac{A(x - 1)(x^2 + 1)}{(x - 1)^2(x^2 + 1)} + \frac{B(x^2 + 1)}{(x - 1)^2(x^2 + 1)} + \frac{(Cx + D)(x - 1)^2}{(x - 1)^2(x^2 + 1)}$

   Now, simplify each numerator: $A(x - 1)(x^2 + 1) = A(x^3 - x^2 + x - 1)$ $B(x^2 + 1) = Bx^2 + B$ $(Cx + D)(x - 1)^2 = (Cx + D)(x^2 - 2x + 1) = Cx(x^2 - 2x + 1) + D(x^2 - 2x + 1)$ Simplifying further: $Cx(x^2 - 2x + 1) = Cx^3 - 2x^2 + Cx$ $D(x^2 - 2x + 1) = Dx^2 - 2Dx + D$

   Thus, the total numerator on the right-hand side is: $A(x^3 - x^2 + x - 1) + (Bx^2 + B) + (Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D)$ Combine like terms: $(A + C)x^3 + (-A - 2C + B + D)x^2 + (A + C - 2D)x - (A - B + D)$

3. Equate numerators: The numerator on the left-hand side is simply $2$, and there is no power of $x$. Therefore, all terms involving $x^3$, $x^2$, and $x$ on the right-hand side must vanish. This gives the system of equations: $A + C = 0$ $-A - 2C + B + D = 0$ $A + C - 2D = 0$ $-(A - B + D) = 2$

4. Solve the system: From $A + C = 0$, we get $A = -C$.

   Substitute $A = -C$ into the second equation: $-(-C) - 2C + B + D = 0 \quad \Rightarrow \quad C - 2C + B + D = 0 \quad \Rightarrow \quad -C + B + D = 0 \quad \Rightarrow \quad B = C - D$

   Now substitute $A = -C$ into the third equation: $-C + C - 2D = 0 \quad \Rightarrow \quad -2D = 0 \quad \Rightarrow \quad D = 0$

   With $D = 0$, the equation $B = C - D$ gives $B = C$.

   Finally, substitute $A = -C$ and $D = 0$ into the last equation: $-(-C - B + D) = 2 \quad \Rightarrow \quad C - C + 0 = 2 \quad \Rightarrow \quad C = 2$

   Therefore, $C = 2$, $A = -2$, $B = 2$, and $D = 0$.

Final Values:

$A = -2, \quad B = 2, \quad C = 2, \quad D = 0$

Would you like more details on any step or have any questions?

Related Questions:

1. How can we generalize partial fractions for higher powers of polynomials?
2. What is the method of partial fraction decomposition and when is it used?
3. Can we apply this method to improper fractions? If yes, how?
4. What is the significance of solving such partial fractions in integration?
5. How do the roots of the denominator affect the decomposition?

Tip:

Always simplify the terms in partial fractions carefully to avoid errors in combining like terms.