3.5 Q-14

Question Statement

We are asked to solve the following integral:

$$\int \frac{1}{(x + 1)^2(x - 1)} , dx$$

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Background and Explanation

This problem involves partial fraction decomposition, a method that breaks a rational function into simpler fractions, making it easier to integrate each term separately. Here, the denominator has a repeated factor $(x + 1)^2$ and a linear factor $(x - 1)$, so we decompose the expression into the following form:

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

We will solve for the constants $A$, $B$, and $C$ by equating terms after multiplying through by the common denominator.

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Solution

We begin by expressing the integrand as a sum of partial fractions:

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

Next, we multiply both sides of equation (i) by $(x + 1)^2(x - 1)$ to eliminate the denominators:

$$1 = A(x + 1)^2 + B(x + 1)(x - 1) + C(x - 1) \tag{ii}$$

Step 1: Solve for $A$

Substitute $x = -1$ into equation (ii) to solve for $A$:

$$1 = A(1 + 1)^2 1 = A(2)^2 \quad \Rightarrow \quad 4A = 1 A = \frac{1}{4}$$

Step 2: Solve for $C$

Substitute $x = 1$ into equation (ii) to solve for $C$:

$$1 = C(1 - 1) 1 = C(-2) \quad \Rightarrow \quad -2C = 1 C = \frac{-1}{2}$$

Step 3: Solve for $B$

Now, equate the coefficients of $x^2$ from both sides of equation (ii) to solve for $B$:

$$0 = A + B 0 = \frac{1}{4} + B \quad \Rightarrow \quad B = -\frac{1}{4}$$

Thus, we can rewrite the integrand as:

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

Step 4: Integrate Each Term

Now, we integrate each term:

1. $\int \frac{\frac{1}{4}}{x - 1} , dx = \frac{1}{4} \ln |x - 1|$
2. $\int \frac{\frac{1}{4}}{x + 1} , dx = \frac{1}{4} \ln |x + 1|$
3. $\int \frac{\frac{1}{2}}{(x + 1)^2} , dx = \frac{1}{2} \left( -\frac{1}{x + 1} \right) = -\frac{1}{2(x + 1)}$

Thus, the integral becomes:

$$\frac{1}{4} \ln |x - 1| - \frac{1}{4} \ln |x + 1| - \frac{1}{2(x + 1)} + C$$

Final Answer:

Therefore, the solution to the integral is:

$$\boxed{\frac{1}{4} \ln \left|\frac{x - 1}{x + 1}\right| + \frac{1}{2(x + 1)} + C}$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Used to break the rational function into simpler fractions.
• Integration of Common Rational Functions:
  • $\int \frac{1}{x} , dx = \ln |x|$
  • $\int \frac{1}{x^2} , dx = -\frac{1}{x}$

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Summary of Steps

1. Express the integrand as a sum of partial fractions.
2. Solve for the unknown coefficients $A$, $B$, and $C$ by substituting values for $x$.
3. Rewrite the integral with the solved coefficients.
4. Integrate each term using standard integration rules.
5. Combine the results and simplify the expression.