3.5 Q-20

Question Statement

Evaluate the following integral:

$$\int \frac{9 x - 7}{(x + 3)(x^2 + 1)} , dx$$

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Background and Explanation

This problem involves partial fraction decomposition to break the integrand into simpler terms that are easier to integrate. We will express the rational function as a sum of simpler fractions and then integrate each term individually. To solve this, we need to know how to decompose rational functions and use standard integration formulas like logarithmic and arctangent integrals.

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Solution

We begin by decomposing the given rational function using partial fractions. We assume the decomposition is of the form:

$$\frac{9x - 7}{(x + 3)(x^2 + 1)} = \frac{A}{x + 3} + \frac{B x + C}{x^2 + 1} \tag{1}$$

Step 1: Multiply both sides by $(x + 3)(x^2 + 1)$ to eliminate the denominators:

$$9x - 7 = A(x^2 + 1) + (B x + C)(x + 3) \tag{2}$$

Step 2: Substitute $x = -3$ to solve for $A$:

By plugging $x = -3$ into equation (2):

$$9(-3) - 7 = A((-3)^2 + 1) \quad \Rightarrow \quad -27 - 7 = A(9 + 1)$$

Simplifying:

$$-34 = A(10) \quad \Rightarrow \quad A = \frac{-34}{10} = \frac{-17}{5}$$

Step 3: Compare coefficients of $x^2$ and $x$ on both sides of equation (2):

• For $x^2$: The coefficient on the left-hand side is 0, and on the right-hand side it is $A + B$. Therefore, we have: $0 = A + B \quad \Rightarrow \quad B = -A = \frac{17}{5}$

• For $x$: The coefficient on the left-hand side is 9, and on the right-hand side it is $3B + C$. Therefore, we have: $9 = 3B + C \quad \Rightarrow \quad 9 = 3\left(\frac{17}{5}\right) + C$ Simplifying: $9 = \frac{51}{5} + C \quad \Rightarrow \quad C = 9 - \frac{51}{5} = \frac{-6}{5}$

Step 4: Substitute the values of $A$, $B$, and $C$ into the partial fractions:

Thus, the partial fraction decomposition is:

$$\frac{9x - 7}{(x + 3)(x^2 + 1)} = \frac{-17}{5(x + 3)} + \frac{17x - 6}{5(x^2 + 1)}$$

Step 5: Integrate each term:

Now we integrate both terms:

$$\int \frac{9x - 7}{(x + 3)(x^2 + 1)} , dx = \int \left( \frac{-17}{5(x + 3)} + \frac{17x - 6}{5(x^2 + 1)} \right) , dx$$

• The first integral is straightforward: $\int \frac{1}{x + 3} , dx = \ln |x + 3|$

• For the second integral, split it into two parts: $\int \frac{17x}{x^2 + 1} , dx = \frac{17}{2} \ln(x^2 + 1)$ and $\int \frac{-6}{x^2 + 1} , dx = -6 \tan^{-1}(x)$

Thus, the final solution is:

$$\frac{-17}{5} \ln |x + 3| + \frac{17}{10} \ln(x^2 + 1) - \frac{6}{5} \tan^{-1}(x) + C$$

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Key Formulas or Methods Used

1. Partial Fraction Decomposition: Breaking a complex rational function into simpler fractions to facilitate integration. $\frac{9x - 7}{(x + 3)(x^2 + 1)} = \frac{A}{x + 3} + \frac{B x + C}{x^2 + 1}$

2. Standard Integrals:

   • $\int \frac{1}{x + a} , dx = \ln |x + a|$
   • $\int \frac{x}{x^2 + a^2} , dx = \frac{1}{2} \ln(x^2 + a^2)$
   • $\int \frac{1}{x^2 + a^2} , dx = \tan^{-1}(x)$

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Summary of Steps

1. Decompose the integrand using partial fractions: $\frac{9x - 7}{(x + 3)(x^2 + 1)} = \frac{-17}{5(x + 3)} + \frac{17x - 6}{5(x^2 + 1)}$

2. Solve for constants $A$, $B$, and $C$ using substitution and coefficient comparison.

3. Integrate each term:

   • Integrate $\frac{1}{x + 3}$ to get $\ln |x + 3|$.
   • Split and integrate $\frac{17x}{x^2 + 1}$ and $\frac{-6}{x^2 + 1}$ to get $\frac{17}{10} \ln(x^2 + 1)$ and $-\frac{6}{5} \tan^{-1}(x)$.

4. Combine the results to obtain the final solution.