If you know that $$\int_{0}^{\infty} e^{-\alpha x^{2}} dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$$ then by differentiating twice under the integral sign, you have:
$$\frac{\partial^{2}}{\partial \alpha^{2}} \int_{0}^{\infty} e^{-\alpha x^{2}} dx = \int_{0}^{\infty} x^{4} e^{-\alpha x^{2}} dx$$
Therefore, applying the differentiation to the result you already know:
$$\frac{\partial^{2}}{\partial \alpha^{2}} \int_{0}^{\infty} e^{-\alpha x^{2}} dx = \frac{\partial^{2}}{\partial \alpha^{2}} \frac{1}{2}\sqrt{\frac{\pi}{\alpha}} = \frac{3\sqrt{\pi}}{8\alpha^{5/2}}$$
In your problem, simply substitute $\alpha = 1/\beta^{2}$ and you have the result that you can verify in W|A:
$$\int_{0}^{\infty} x^{4} e^{-\alpha x^{2}} dx = \frac{3\sqrt{\pi}}{8\left(\frac{1}{\beta^{2}}\right)^{5/2}}$$
Given your recent update, let me add the following:
You should know that the expectation and variance are given by
$$E[X] = \int x f(x)dx$$
and $$Var(X) = \int (x-\mu)^2f(x)dx$$
I'm sure you can integrate by substitution the first one. The variance is a bit easier once you know how to apply the Gaussian integral above, since it involves a term $\int x^{2} f(x) dx$ and a constant $\mu$. In that case, of course, you only need to differentiate once.