Try A Substitution First

$\begin{aligned} I & = \int_0^\infty x^2 e^{-x^2} dx \quad \quad \small \color{#3D99F6}{\text{Let }t = x^2 \quad \Rightarrow dx = \frac{1}{2\sqrt{t}}dt} \\ & = \int_0^\infty \frac{t e^{-t}}{2\sqrt{t}} dx \\ & = \frac{1}{2} \int_0^\infty t^\frac{1}{2} e^{-t} dx \\ & = \frac{1}{2} \int_0^\infty t^{\color{#3D99F6}{\frac{3}{2}}-1} e^{-t} dx \\ & = \frac{1}{2} \color{#3D99F6}{\Gamma \left(\frac{3}{2}\right)} \quad \quad \small \color{#3D99F6}{\Gamma (x) \text{ is Gamma function}} \\ & = \frac{1}{2} \cdot{} \frac{1}{2} \Gamma \left(\frac{1}{2}\right) \\ & = \boxed{\dfrac{\sqrt{\pi}}{4}} \end{aligned}$

$\begin{aligned} \int_{0}^{\infty} x^2e^{-x^2} &=\frac{1}{4}(\ce{erf}(x)-2e^{-x^2}x) \quad\quad\quad\quad\quad\quad{\text{Use }\int_{a}^{b} f(x) dx = F(b) - F(a) \space \ce{or} \lim_{x \to b^-}(F(x)) -\lim_{x \to a^+}(F(x)) } \\&\therefore \frac{\pi}{4} \space \square \end{aligned}$

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$\ce{erf}(x)$ is an error function

$\LARGE \text{ADIOS!!! Muchacha}$