It seems you're asking to compute the integral:

\[
\int \frac{6 \, dx}{1 + x^2}
\]

This integral can be solved using a standard formula from calculus:

\[
\int \frac{1 \, dx}{1 + x^2} = \tan^{-1}(x) + C
\]

Applying this to your integral:

\[
\int \frac{6 \, dx}{1 + x^2} = 6 \cdot \tan^{-1}(x) + C
\]

where \( C \) is the constant of integration.

Would you like a detailed explanation or have any further questions?

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Here are 5 related questions you might find interesting:

1. What is the integral of \( \frac{1}{1 + x^2} \) from 0 to 1?
2. How do you derive the formula for the inverse tangent function from its derivative?
3. Can you integrate \( \frac{1}{1 + x^2} \) using a trigonometric substitution?
4. What are the properties of the arctangent function, \(\tan^{-1}(x)\)?
5. How would you compute the derivative of \( \tan^{-1}(x) \)?

**Tip:** Remember that integrals involving \( \frac{1}{1 + x^2} \) often lead to inverse trigonometric functions, like \( \tan^{-1}(x) \).

Learn how to solve the integral of 6/(1 + x^2) with a step-by-step explanation. This problem uses inverse trigonometric functions and is suitable for students in Grades 11-12. Understand the key formula ∫ 1 / (1 + x^2) dx = tan^(-1)(x) + C, and how it applies to this problem.

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