The integral shown in the image is:
$\int_{0}^{\infty} \frac{2x}{1 + x^4} \, dx$

Solution:

To evaluate this improper integral, we can proceed as follows:

1. Substitution:
   Let $u = x^2$, which implies $du = 2x \, dx$. This substitution simplifies the integral: $\int_{0}^{\infty} \frac{2x}{1 + x^4} \, dx = \int_{0}^{\infty} \frac{du}{1 + u^2}.$

2. Recognizing the standard arctangent integral form:
   The integral $\int \frac{1}{1 + u^2} \, du$ is the standard formula for $\arctan(u) + C$.

3. Apply the limits:
   $\int_{0}^{\infty} \frac{du}{1 + u^2} = \left[ \arctan(u) \right]_{0}^{\infty}.$ Evaluate:

   • At $u = \infty$: $\arctan(\infty) = \frac{\pi}{2}$,
   • At $u = 0$: $\arctan(0) = 0$.

4. Final result:
   $\int_{0}^{\infty} \frac{2x}{1 + x^4} \, dx = \frac{\pi}{2} - 0 = \frac{\pi}{2}.$

Conclusion:

$\boxed{\frac{\pi}{2}}$

Would you like a more detailed explanation or further examples?

Here are 5 related questions to deepen understanding:

1. How does substitution help simplify integrals with polynomial denominators?
2. What is the integral of $\frac{1}{1 + x^2}$ and how is it related to trigonometric functions?
3. How would the result change if the bounds of integration were finite instead of infinite?
4. Can improper integrals always be evaluated using substitution?
5. How does the convergence of an improper integral depend on the behavior of the integrand?

Tip: Recognizing standard integral forms like $\int \frac{1}{1 + x^2} \, dx = \arctan(x) + C$ is crucial for solving many calculus problems efficiently!