To solve the integral $\int \frac{11x^2 \, dx}{(81 + x^2)^2}$, we can use trigonometric substitution.

Part 1: Choosing the Appropriate Trigonometric Substitution

The expression $81 + x^2$ suggests a trigonometric substitution that matches the form $a^2 + x^2$, where $a = 9$. In such cases, the appropriate substitution is:

$x = 9 \tan(\theta)$

This is because $\tan(\theta)$ works well with integrals of the form $a^2 + x^2$, simplifying the square terms.

Thus, the correct substitution is Option B: $x = 9 \tan(\theta)$.

Part 2: Rewrite the Integral Using the Substitution

With $x = 9 \tan(\theta)$, we now need to compute the following:

• Derivative: $dx = 9 \sec^2(\theta) \, d\theta$
• $x^2 = 81 \tan^2(\theta)$
• $81 + x^2 = 81 + 81 \tan^2(\theta) = 81 \sec^2(\theta)$

Now, rewrite the integral:

$\int \frac{11x^2 \, dx}{(81 + x^2)^2} = \int \frac{11(81 \tan^2(\theta)) \cdot 9 \sec^2(\theta) \, d\theta}{(81 \sec^2(\theta))^2}$

Simplifying the integral:

$= \int \frac{11 \cdot 81 \tan^2(\theta) \cdot 9 \sec^2(\theta) \, d\theta}{81^2 \sec^4(\theta)}$

$= \int \frac{11 \cdot 9 \tan^2(\theta)}{81 \sec^2(\theta)} \, d\theta$

$= \int \frac{99 \tan^2(\theta)}{81 \sec^2(\theta)} \, d\theta$

$= \frac{99}{81} \int \frac{\tan^2(\theta)}{\sec^2(\theta)} \, d\theta$

Since $\frac{\tan^2(\theta)}{\sec^2(\theta)} = \sin^2(\theta)$, we now have:

$\frac{99}{81} \int \sin^2(\theta) \, d\theta$

Use the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$:

$\frac{99}{81} \int \frac{1 - \cos(2\theta)}{2} \, d\theta$

This becomes:

$\frac{99}{81} \cdot \frac{1}{2} \int (1 - \cos(2\theta)) \, d\theta$

$= \frac{99}{162} \left( \theta - \frac{\sin(2\theta)}{2} \right)$

Back-Substitution

To finish the evaluation, we need to revert the substitution $x = 9 \tan(\theta)$. From $x = 9 \tan(\theta)$, we have $\theta = \tan^{-1}\left(\frac{x}{9}\right)$. Also, using the identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we can express $\sin(2\theta)$ in terms of $x$.

Thus, the evaluated integral is:

$\frac{99}{162} \left( \tan^{-1}\left(\frac{x}{9}\right) - \frac{x}{81 + x^2} \right) + C$

Final Answer:

$\frac{99}{162} \left( \tan^{-1}\left(\frac{x}{9}\right) - \frac{x}{81 + x^2} \right) + C$

Would you like more details on any specific part of this solution? Here are five related questions:

1. How do we choose trigonometric substitutions for different types of integrals?
2. What is the derivation of the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$?
3. How do we handle back-substitution in trigonometric substitution problems?
4. Can trigonometric substitutions be used for other integrals, such as those involving $\sqrt{x^2 - a^2}$?
5. Why does the integral simplify when using trigonometric identities?

Tip: When performing trigonometric substitution, always check if the final result can be simplified further using trigonometric identities or algebraic manipulation.