To solve the integral

$\int \frac{1}{1+x^2} \, dx$

using the substitution $x = \tan u$, we follow these steps:

1. Substitution: If $x = \tan u$, then $dx = \sec^2 u \, du$.

2. Change the integral: Substituting $x$ and $dx$ into the integral gives:

   $\int \frac{1}{1+\tan^2 u} \sec^2 u \, du$

3. Simplify using the identity: Recall the trigonometric identity $1 + \tan^2 u = \sec^2 u$. Thus, the integral simplifies to:

   $\int \frac{\sec^2 u}{\sec^2 u} \, du = \int 1 \, du$

4. Integrate: The integral of $1$ with respect to $u$ is:

   $u + C$

5. Back-substitution: Since $u = \tan^{-1}(x)$, we have:

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

Thus, the final answer is:

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

Questions

1. Would you like to see a graphical representation of the integral?
2. Do you want to explore other substitution methods for this integral?
3. Would you like to see the application of this integral in real-world problems?
4. Do you want to learn about the properties of the arctangent function?
5. Would you like to practice similar integrals?

Tip

When working with trigonometric substitutions, it's helpful to remember the identities associated with them, as they can greatly simplify the integration process.