I was watching this: https://youtu.be/qQ-56b_LvOw?t=4484
And this integral came up. $$\int_{0}^{\infty}{(2x^2+1)e^{-x^2}}dx$$
To which the answer was $\sqrt{\pi}$.
They made it clear that you didn't need knowledge about erf$(x)$ to solve this integral, although it looks like one of the competitors tried to solve it using that. How can you solve this without "extra knowledge"?(Assuming you only have mastered the techniques of integral calculus)
We need to first compute the integral
$$ I = \int_0^{\infty} e^{-x^2}\,dx.$$
Note that
$$I^2 = \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2 = \int_0^{\infty} e^{-x^2}\,dx\int_0^{\infty} e^{-y^2}\,dy.$$
I used a different variable for the second integral since the two integrals would get confused if we used the same integration variable for each.
Recombining our integrals, we get
$$ I^2 = \int_0^{\infty}\int_0^{\infty} e^{-x^2-y^2}\,dxdy.$$
Let us now switch to polar coordinates so that $x^2 + y^2 = r^2$ and $dxdy = r\,drd\theta$. In the $xy$ plane we are integrating over the first quadrant which corresponds to $r$ ranging from $0$ to $\infty$ and $\theta$ ranging from $0$ to $\frac{\pi}{2}$. We then have
$$ I^2 = \int_0^{\frac{\pi}{2}}\int_0^{\infty} e^{-r^2}r\,drd\theta.$$
Letting $z = r^2$, we have $dz = 2r\,dr$ and so
$$ I^2 = \int_0^{\frac{\pi}{2}} \int_0^{\infty} e^{-z}\frac{1}{2}dzd\theta.$$
We can do the $z$ integral very easily now: $\int_0^{\infty} e^{-z}\,dz = 1$, giving
$$ I^2 = \frac{1}{2}\int_0^{\frac{\pi}{2}} d\theta = \frac{\pi}{4}.$$
Taking a square root, we see that $I = \frac{\sqrt\pi}{2}$. This accounts only for the second half of the integral. The first half will have to be treated in a different manner.
Consider then
$$ J(\alpha) = \int_0^{\infty} e^{-\alpha x^2}\,dx.$$
Note that if we take an $\alpha$ derivative of both sides, we have
$$ J'(\alpha) = \frac{d}{d\alpha} \int_0^{\infty} e^{-\alpha x^2}\,dx = \int_0^{\infty} (-x^2)e^{-\alpha x^2}\,dx.$$
The way to think about this is that with respect to $\alpha$, $x$ is constant so if we differentiate the integrand $e^{-\alpha x^2}$ with respect to $\alpha$, the $-x^2$ just comes out front via chain rule. So then if we plug in $\alpha=1$, we have that
$$ J'(1) = -\int_0^{\infty} x^2 e^{-\alpha x^2}\,dx$$
which is quite nearly exactly what we wanted (modulo a factor of $-2$). This means that if we can evaluate $J(\alpha)$ in terms of $\alpha$, we would be done. Let us do so. Setting $z = \sqrt{\alpha} x$, $dz = \sqrt{\alpha}\,dx$, we get
$$ J(\alpha) = \int_0^{\infty} e^{-\alpha x^2}\,dx = \int_0^{\infty} e^{-z^2}\,\frac{dz}{\sqrt{\alpha}}.$$
We already computed the integral above, so we get that
$$ J(\alpha) = \frac{\sqrt{\pi}}{2\sqrt{\alpha}}$$
and so
$$ J'(\alpha) = -\frac{\sqrt{\pi}}{4\alpha^{\frac{3}{2}}}.$$
Plugging in $\alpha = 1$ and doing some minor adjustments, you can get your answer.
Using integration by parts we get $$ \begin{align} \int\limits_0^\infty (2x^2 + 1) e^{-x^2} \, dx &= \int\limits_0^\infty \underbrace{x}_f (\underbrace{2x e^{-x^2}}_{g'}) \, dx + \int\limits_0^\infty e^{-x^2} \, dx \\ &= \left[\underbrace{x}_f \, (\underbrace{-e^{-x^2}}_g) \right]_{x=0}^{x\to\infty} - \int\limits_0^\infty(\underbrace{1}_{f'})(\underbrace{-e^{-x^2}}_g) \, dx + \int\limits_0^\infty e^{-x^2} \, dx \\ &= 2 \int\limits_0^\infty e^{-x^2} \, dx \\ &= \int\limits_{-\infty}^\infty e^{-x^2} \, dx \\ &= I \end{align} $$ To solve $I = \sqrt{\pi}$ one uses the trick by Poisson (solving $I^2$ as integral over the plane in polar coordinates), as answered many times on this site already, e.g. here.
