The given integral is:

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

We are asked to solve this using trigonometric substitution.

Step-by-Step Solution:

Step 1: Completing the square

The expression inside the square root $x^2 + x + 1$ can be rewritten by completing the square:

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

So, the integral becomes:

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

Step 2: Trigonometric substitution

Next, we use the substitution for this type of expression. Set:

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

Thus, differentiating both sides:

$dx = \frac{\sqrt{3}}{2} \sec^2(\theta) d\theta$

Substitute into the integral:

$I = \int \left( \frac{\sqrt{3}}{2} \tan(\theta) - \frac{1}{2} \right) \left( \left( \frac{\sqrt{3}}{2} \tan(\theta) \right)^2 + \frac{3}{4} \right)^{-\frac{1}{2}} \cdot \frac{\sqrt{3}}{2} \sec^2(\theta) d\theta$

Simplify the expression under the square root:

$\left( \frac{\sqrt{3}}{2} \tan(\theta) \right)^2 + \frac{3}{4} = \frac{3}{4} \left( \tan^2(\theta) + 1 \right) = \frac{3}{4} \sec^2(\theta)$

So the integral becomes:

$I = \int \left( \frac{\sqrt{3}}{2} \tan(\theta) - \frac{1}{2} \right) \cdot \left( \frac{3}{4} \sec^2(\theta) \right)^{-\frac{1}{2}} \cdot \frac{\sqrt{3}}{2} \sec^2(\theta) d\theta$

Since $\left( \frac{3}{4} \sec^2(\theta) \right)^{-\frac{1}{2}} = \frac{2}{\sqrt{3}} \cdot \frac{1}{\sec(\theta)}$, the integral simplifies further:

$I = \int \left( \frac{\sqrt{3}}{2} \tan(\theta) - \frac{1}{2} \right) \sec(\theta) d\theta$

Step 3: Break into two integrals

Now, distribute and break the integral into two parts:

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

• The first integral $\int \tan(\theta) \sec(\theta) d\theta$ is straightforward:

$\int \tan(\theta) \sec(\theta) d\theta = \sec(\theta)$

• The second integral $\int \sec(\theta) d\theta$ is a standard result:

$\int \sec(\theta) d\theta = \ln |\sec(\theta) + \tan(\theta)|$

Step 4: Substitution back

Now, substituting back $\theta$ in terms of $x$:

Recall $\tan(\theta) = \frac{2}{\sqrt{3}} \left( x + \frac{1}{2} \right)$. Therefore:

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

Thus, the final solution is a combination of these results:

$I = \frac{\sqrt{3}}{2} \sec(\theta) - \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C$

This is the final answer. Would you like further explanation on any specific part of this process?

Related questions:

1. How do you choose the right trigonometric substitution for an integral?
2. What are the most common trigonometric substitutions for different forms?
3. How does completing the square help simplify integrals?
4. Can you explain how to simplify expressions like $\sec^2(\theta)$ after substitution?
5. How do you revert back to $x$ after integrating in terms of $\theta$?

Tip:

Always remember to use the correct trigonometric identity when simplifying the integrand after substitution!