3.5 Q-12

Question Statement

We are asked to solve the following integral:

$$\int \frac{4 + 7x}{(1 + x)^2(2 + 3x)} , dx$$

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Background and Explanation

This problem requires us to solve a rational function using partial fraction decomposition. Partial fractions help break down a complex rational function into simpler fractions that can be integrated individually. To apply this technique, we will express the given rational function as a sum of simpler fractions, each corresponding to the factors in the denominator.

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Solution

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

$$\frac{4 + 7x}{(1 + x)^2(2 + 3x)} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2} + \frac{C}{2 + 3x} \tag{i}$$

Now, multiply both sides of equation (i) by the denominator $(1 + x)^2(2 + 3x)$ to eliminate the denominators:

$$4 + 7x = A(1 + x)(2 + 3x) + B(2 + 3x) + C(1 + x)^2 \tag{ii}$$

Step 1: Solve for $B$

Substitute $x = -1$ into equation (ii) to find $B$:

$$4 + 7(-1) = B(2 + 3(-1)) 4 - 7 = B(2 - 3) -3 = -B \Rightarrow B = 3$$

Step 2: Solve for $C$

Substitute $x = \frac{-2}{3}$ into equation (ii) to find $C$:

$$4 + 7\left(\frac{-2}{3}\right) = C\left(1 - \frac{-2}{3}\right)^2 \frac{12 - 14}{3} = C\left(\frac{3 - 2}{3}\right)^2 \frac{-2}{3} = C\left(\frac{1}{9}\right) C = \frac{-2}{3} \times \frac{9}{1} = -6$$

Step 3: Solve for $A$

To find $A$, we equate the coefficients of like terms from both sides of equation (ii):

$$4 = 2A + 2B + C 4 = 2A + 2(3) - 6 4 = 2A + 6 - 6 A = 2$$

Thus, we can rewrite the original integral as:

$$\int \frac{4 + 7x}{(1 + x)^2(2 + 3x)} , dx = \int \left[ \frac{2}{1 + x} + \frac{3}{(1 + x)^2} + \frac{-6}{2 + 3x} \right] , dx$$

Step 4: Integrate the terms

Now we can integrate each term individually:

1. The first term: $\int \frac{2}{1 + x} , dx = 2 \ln |1 + x|$
2. The second term: $\int \frac{3}{(1 + x)^2} , dx = -\frac{3}{1 + x}$
3. The third term: $\int \frac{-6}{2 + 3x} , dx$ involves a substitution. Let $u = 2 + 3x$, so $du = 3dx$, and we get: $\int \frac{-6}{2 + 3x} , dx = -2 \ln |2 + 3x|$

Therefore, the integral is:

$$2 \ln |1 + x| - \frac{3}{1 + x} - 2 \ln |2 + 3x| + C$$

Final Answer:

Thus, the solution to the integral is:

$$\boxed{\ln(1 + x)^2 - \ln(2 + 3x)^2 - \frac{3}{1 + x} + C}$$

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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:
  • $\int \frac{1}{x} , dx = \ln |x|$
  • $\int \frac{1}{x^2} , dx = -\frac{1}{x}$
  • Substitution for linear functions in the denominator.

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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.