The integral presented in the image is:

$\int \frac{\left( \tan^{-1}x \right)^2 dx}{x^2 + 1}, \quad u = \tan^{-1}x$

To solve this, let's use the substitution provided:

Step-by-Step Solution:

1. Substitute $u = \tan^{-1}x$: $u = \tan^{-1}x \implies \frac{du}{dx} = \frac{1}{x^2 + 1}$ Therefore, we can replace $\frac{dx}{x^2 + 1}$ with $du$.

2. Rewrite the Integral: Using the substitution $u = \tan^{-1}x$, the given integral becomes: $\int u^2 du$

3. Integrate: The integral of $u^2$ is a standard power rule integration: $\int u^2 du = \frac{u^3}{3} + C$

4. Substitute $u = \tan^{-1}x$ back: Now, replace $u$ with $\tan^{-1}x$: $\frac{(\tan^{-1}x)^3}{3} + C$

Final Answer:

$\boxed{\frac{(\tan^{-1}x)^3}{3} + C}$

Would you like further clarification or more details on any of the steps?

Here are five follow-up questions you might explore based on this integral:

1. How does the substitution method help simplify more complex integrals?
2. Can you apply a similar substitution method to integrals involving $\sin^{-1}x$ or $\cos^{-1}x$?
3. What are other possible methods to solve integrals involving inverse trigonometric functions?
4. How does the integration of powers of inverse trigonometric functions differ from the integration of trigonometric functions?
5. Could we use integration by parts for this problem, and how would the approach change?

Tip: When dealing with inverse trigonometric functions, it's often helpful to recognize patterns for substitution early on to simplify the integrals.