3.3 Q-2

Question Statement

Evaluate the integral:

$$\int \frac{dx}{x^2 + 4x + 13}$$

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Background and Explanation

In this problem, we are asked to integrate a rational function where the denominator is a quadratic expression. To solve such integrals, it is often helpful to complete the square in the denominator, transforming it into a form that resembles the standard integral of the form:

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

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Solution

1. Rewrite the denominator:
   Start by completing the square for the quadratic expression in the denominator:

$$x^2 + 4x + 13$$

First, focus on the $x^2 + 4x$ part. To complete the square, take half of the coefficient of $x$, square it, and add and subtract this value. Half of $4$ is $2$, and $2^2 = 4$. So, we rewrite the quadratic as:

$$x^2 + 4x = (x + 2)^2 - 4$$

Now, include the constant term $13$:

$$x^2 + 4x + 13 = (x + 2)^2 + 9$$

So, the integral becomes:

$$\int \frac{dx}{(x + 2)^2 + 3^2}$$

2. Recognize the standard integral form:
   This now matches the standard form for an arctangent integral:

$$\int \frac{dx}{(x + 2)^2 + 3^2} = \frac{1}{3} \tan^{-1}\left(\frac{x + 2}{3}\right) + C$$

Therefore, the result of the integral is:

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

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Key Formulas or Methods Used

• Completing the Square: To rewrite the quadratic expression in a form suitable for applying the standard integral formula.
• Standard Integral: Used the formula for integrating expressions of the form $\frac{1}{x^2 + a^2}$, which results in $\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

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Summary of Steps

1. Complete the square in the denominator:
   $x^2 + 4x + 13 = (x + 2)^2 + 9$

2. Recognize the integral as a standard arctangent form: $\int \frac{dx}{(x + 2)^2 + 3^2} = \frac{1}{3} \tan^{-1}\left(\frac{x + 2}{3}\right) + C$

3. Final result:
   $\frac{1}{3} \tan^{-1}\left(\frac{x + 2}{3}\right) + C$