Let's evaluate these integrals using trigonometric substitution:

(a) $\int \frac{\sqrt{x^2 + 4x + 3}}{x + 2} \, dx$

We begin by simplifying the square root expression. Complete the square inside the square root:

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

Thus, the integral becomes:

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

Now, let’s use the substitution $x + 2 = \sec(\theta)$, which implies $dx = \sec(\theta)\tan(\theta)\, d\theta$. Substituting this into the integral:

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

Solve this using the identity $\tan^2(\theta) = \sec^2(\theta) - 1$.

(b) $\int \frac{dx}{x^3\sqrt{x^2 - 4}}$

Here, we can use the substitution $x = 2\sec(\theta)$, so that $dx = 2\sec(\theta)\tan(\theta)\, d\theta$ and $\sqrt{x^2 - 4} = 2\tan(\theta)$. Substituting into the integral:

$\int \frac{2\sec(\theta)\tan(\theta)}{8\sec^3(\theta)\tan(\theta)} \, d\theta = \frac{1}{4} \int \frac{d\theta}{\sec^2(\theta)}$

Since $\frac{1}{\sec^2(\theta)} = \cos^2(\theta)$, integrate to get:

$\frac{1}{4} \int \cos(\theta)\, d\theta$

(c) $\int \frac{dx}{(x^2 - 2x + 5)^2}$

Complete the square for the quadratic:

$x^2 - 2x + 5 = (x - 1)^2 + 4$

Now, use the substitution $x - 1 = 2\tan(\theta)$, so $dx = 2\sec^2(\theta)\, d\theta$. This simplifies the integral to:

$\int \frac{2\sec^2(\theta)}{16\sec^4(\theta)} \, d\theta = \frac{1}{8} \int \cos^2(\theta) \, d\theta$

Solve using standard trigonometric identities.

(d) $\int \frac{x\, dx}{(9 - x^2)^{3/2}}$

Use the substitution $x = 3\sin(\theta)$, so that $dx = 3\cos(\theta)\, d\theta$ and $9 - x^2 = 9\cos^2(\theta)$. Substituting into the integral:

$\int \frac{3\sin(\theta) \cdot 3\cos(\theta)\, d\theta}{(9\cos^2(\theta))^{3/2}} = \int \frac{9\sin(\theta)\cos(\theta)\, d\theta}{27\cos^3(\theta)} = \frac{1}{3} \int \tan(\theta) \, d\theta$

Would you like further details on these steps? Here are some related questions to deepen your understanding:

1. What is the general approach for completing the square in integrals?
2. How does trigonometric substitution simplify integrals involving square roots?
3. Can trigonometric identities be used to simplify any integral?
4. How do you recognize when to use trigonometric substitution?
5. What are common challenges in solving integrals with trigonometric substitution?

Tip: Always check for opportunities to simplify the integrand before applying substitutions!