The integral in the image is:

$\int_{0}^{\infty} \frac{x \, dx}{1 + x^6}.$

We can solve this integral using the Gamma function and symmetry properties.

Step-by-Step Solution:

1. Substitute: Let $u = x^6$, then $x = u^{1/6}$ and $dx = \frac{1}{6} u^{-5/6} \, du$. Substituting into the integral:

   $\int_{0}^{\infty} \frac{x \, dx}{1 + x^6} = \int_{0}^{\infty} \frac{u^{1/6} \cdot \frac{1}{6} u^{-5/6}}{1 + u} \, du.$

   Simplify the powers of $u$:

   $\int_{0}^{\infty} \frac{x \, dx}{1 + x^6} = \frac{1}{6} \int_{0}^{\infty} \frac{u^{-2/3}}{1 + u} \, du.$

2. Use the Beta Function Relation:

   The integral $\int_{0}^{\infty} \frac{u^a}{(1+u)^b} \, du$ can be expressed in terms of the Beta function:

   $\int_{0}^{\infty} \frac{u^{a-1}}{(1+u)^b} \, du = B(a, b-a),$

   where the Beta function $B(x, y)$ relates to the Gamma function as:

   $B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}.$

   Here, $a = \frac{1}{3}$ and $b = 1$. Thus, the integral becomes:

   $\frac{1}{6} \int_{0}^{\infty} \frac{u^{-2/3}}{1+u} \, du = \frac{1}{6} B\left(\frac{1}{3}, 1 - \frac{1}{3}\right).$

3. Simplify the Beta Function:

   Substitute into the Beta function formula:

   $B\left(\frac{1}{3}, \frac{2}{3}\right) = \frac{\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right)}{\Gamma(1)}.$

   Using the property $\Gamma(1) = 1$ and the reflection formula for the Gamma function:

   $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)},$

   we get:

   $\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right) = \frac{\pi}{\sin\left(\frac{\pi}{3}\right)}.$

   Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, this becomes:

   $\Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right) = \frac{\pi}{\frac{\sqrt{3}}{2}} = \frac{2\pi}{\sqrt{3}}.$

4. Final Result:

   Substituting back, the value of the integral is:

   $\int_{0}^{\infty} \frac{x \, dx}{1 + x^6} = \frac{1}{6} \cdot B\left(\frac{1}{3}, \frac{2}{3}\right) = \frac{1}{6} \cdot \frac{2\pi}{\sqrt{3}} = \frac{\pi}{3\sqrt{3}}.$

Final Answer:

$\int_{0}^{\infty} \frac{x \, dx}{1 + x^6} = \frac{\pi}{3\sqrt{3}}.$

Would you like more details or related concepts explained?
Here are 5 related questions to explore:

1. How is the Beta function related to the Gamma function?
2. Can we solve this integral using residue calculus?
3. What is the reflection formula for the Gamma function and why is it useful?
4. How do symmetry properties simplify improper integrals?
5. How would this integral change if the power of $x^6$ were different?

Tip: Remember, the Gamma and Beta functions are powerful tools for solving complex integrals, especially when the limits involve infinity!