3.3 Q-17

Question Statement

Evaluate the following integral:

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

---

Background and Explanation

To solve this integral, we will need to apply a series of substitutions, including completing the square and trigonometric substitution. Understanding these techniques will help simplify the given rational expression into a solvable form.

---

Solution

Let’s go step-by-step through the solution:

1. Rewrite the Denominator: Start by rewriting the denominator in a more manageable form:

$$4 + 2x + x^2 = (x^2 + 2x + 1) + 3 = (x + 1)^2 + 3$$

The integral now becomes:

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

2. Substitute for $x + 1$: Let:

$$x + 1 = \sqrt{3} \tan \theta$$

This implies:

$$dx = \sqrt{3} \sec^2 \theta , d\theta$$

Substituting this into the integral gives:

$$\int \frac{\left(\sqrt{3} \tan \theta - 1\right) \sqrt{3} \sec^2 \theta , d\theta}{3 (\tan^2 \theta + 1)}$$

3. Simplify the Expression: Recall that $\tan^2 \theta + 1 = \sec^2 \theta$, so the integral simplifies to:

$$\int \frac{(\sqrt{3} \tan \theta - 1) \sqrt{3} \sec^2 \theta , d\theta}{3 \sec^2 \theta}$$

The $\sec^2 \theta$ terms cancel, and we are left with:

$$\frac{1}{\sqrt{3}} \int (\sqrt{3} \tan \theta - 1) , d\theta$$

4. Integrate: Now, we can split the integral:

$$\frac{1}{\sqrt{3}} \left( \int \sqrt{3} \tan \theta , d\theta - \int 1 , d\theta \right)$$

The integral of $\tan \theta$ is $\ln |\sec \theta|$, and the integral of 1 is $\theta$. This gives:

$$\frac{1}{\sqrt{3}} \left( \sqrt{3} \ln |\sec \theta| - \theta \right) + C$$

5. Substitute Back $\theta$: We need to substitute back for $\theta$. Since:

$$x + 1 = \sqrt{3} \tan \theta$$

we have:

$$\tan \theta = \frac{x + 1}{\sqrt{3}}, \quad \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + \frac{(x + 1)^2}{3}} = \frac{\sqrt{3 + (x + 1)^2}}{\sqrt{3}}$$

Substituting these into the expression:

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

6. Final Result: Simplifying further, we obtain:

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

---

Key Formulas or Methods Used

• Completing the Square: Used to rewrite the denominator in the form $(x + 1)^2 + 3$.
• Trigonometric Substitution: $x + 1 = \sqrt{3} \tan \theta$ to simplify the integral.
• Standard Integrals:
  • $\int \tan \theta , d\theta = \ln |\sec \theta|$
  • $\int 1 , d\theta = \theta$

---

Summary of Steps

1. Rewrite the denominator as $(x + 1)^2 + 3$.
2. Use substitution $x + 1 = \sqrt{3} \tan \theta$ and simplify the integral.
3. Perform the integration, splitting the terms.
4. Substitute back $\theta$ using $\tan \theta = \frac{x + 1}{\sqrt{3}}$ and $\sec \theta = \frac{\sqrt{3 + (x + 1)^2}}{\sqrt{3}}$.
5. Simplify the result to obtain the final answer:

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