3.5 Q-21

Question Statement

Evaluate the integral:

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

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Background and Explanation

To solve this problem, we use partial fraction decomposition, which involves expressing the rational function as a sum of simpler fractions. The method allows us to break down the integrand into terms that are easier to integrate. Specifically, we split the rational function into two fractions:

1. One fraction with the denominator $(x-3)$.
2. Another fraction with the denominator $(x^2 + 4)$.

This approach requires determining appropriate constants (A, B, C) to fit the given function.

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Solution

We begin by setting up the partial fraction decomposition. The given rational function is:

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

Step 1: Multiply through by the common denominator

Multiply both sides by $(x-3)(x^2+4)$ to eliminate the denominators:

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

Step 2: Substitute $x = 3$

To solve for $A$, substitute $x = 3$ into equation (ii), which simplifies the equation:

$$1 + 4(3) = A(3^2 + 4) + 0 \quad \Rightarrow \quad 13 = A(9 + 4) \quad \Rightarrow \quad 13 = 13A \quad \Rightarrow \quad A = 1$$

Step 3: Compare coefficients

Now, we equate the coefficients of $x^2$, $x$, and the constant term on both sides of equation (ii).

• Coefficient of $x^2$: $0 = A + B \quad \Rightarrow \quad B = -A = -1$.
• Coefficient of $x$: $4 = -3B + C \quad \Rightarrow \quad C = 4 + 3B = 4 + 3(-1) = 1$.

Step 4: Rewrite the integral

We now substitute the values of $A$, $B$, and $C$ into the partial fraction decomposition:

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

The integral becomes:

$$\int \frac{1}{x-3} , dx + \int \frac{-x + 1}{x^2 + 4} , dx$$

Step 5: Solve each integral

Now, we integrate the two parts separately:

• The first integral is straightforward:

$$\int \frac{1}{x-3} , dx = \ln |x-3|$$

• For the second integral, split it into two parts:

$$\int \frac{-x + 1}{x^2 + 4} , dx = - \int \frac{x}{x^2 + 4} , dx + \int \frac{1}{x^2 + 4} , dx$$

• The first part can be solved using a simple substitution ($u = x^2 + 4$, so $du = 2x , dx$):

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

• The second part is a standard arctangent integral:

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

Step 6: Combine all parts

Finally, combining all the results, we get the solution:

$$\ln |x-3| - \frac{1}{2} \ln(x^2 + 4) + \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$$

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Key Formulas or Methods Used

• Partial Fraction Decomposition: Splitting a rational function into simpler fractions for easier integration.
• Basic Integrals:
  • $\int \frac{1}{x-a} , dx = \ln|x-a|$
  • $\int \frac{1}{x^2 + a^2} , dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$

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Summary of Steps

1. Set up the partial fraction decomposition:
   $\frac{1+4x}{(x-3)(x^2+4)} = \frac{A}{x-3} + \frac{B x + C}{x^2 + 4}$

2. Multiply both sides by the common denominator $(x-3)(x^2+4)$ and simplify.

3. Substitute $x = 3$ to find $A$.

4. Compare the coefficients of $x^2$, $x$, and the constant term to find $B$ and $C$.

5. Rewrite the integral using the values of $A$, $B$, and $C$: $\int \frac{1}{x-3} , dx + \int \frac{-x + 1}{x^2 + 4} , dx$

6. Solve each integral:

   • The first is $\ln |x-3|$.
   • The second splits into two parts: $-\frac{1}{2} \ln(x^2 + 4)$ and $\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right)$.

7. Combine all the results for the final answer.