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I used the result $$frac2pi mathrmexp(-z^2) intlimits_0^infty mathrmexp(-z^2x^2) frac1x^2+1 mathrmdx = mathrmerfc(z)$$ to answer this MSE question. As I mentioned in the link, I obtained this result from the DLMF. I happened to find this solution after failing to evaluate the integral using a variety of substitutions. A solution would be appreciated.

Addendum

Expanding @Jack D'Aurizio's solution, we have

beginalign
frac2pi mathrme^-z^2 intlimits_0^infty fracmathrme^-z^2x^2x^2 + 1 mathrmdx &=
frac2zpi mathrme^-z^2 intlimits_0^infty fracmathrme^-t^2z^2 + t^2 mathrmdt \
&= fraczpi mathrme^-z^2 intlimits_-infty^infty fracmathrme^-t^2z^2 + t^2 mathrmdt
endalign
we used the substitution $x=t/z$.

For the integral
beginequation
intlimits_-infty^infty fracmathrme^-t^2z^2 + t^2 mathrmdt
endequation
we let $f(t) = mathrme^-t^2$ and $g(t) = 1/(z^2 + t^2)$ and take Fourier transforms of each,
beginequation
mathrmF(s) = mathcalF[f(t)] = fracmathrme^-s^2/4sqrt2
endequation
and
beginequation
mathrmG(s) = mathcalF[g(t)] = frac1zsqrtfracpi2 mathrme^-z
endequation
then invoke Parseval's theorem
beginequation
intlimits_-infty^infty f(t)overlineg(t) mathrmdt
= intlimits_-infty^infty mathrmF(s)overlinemathrmG(s) mathrmds
endequation
dropping constants, the integral becomes

beginalign
intlimits_-infty^infty mathrme^-s^2/4 mathrme^-z mathrmds
&= 2intlimits_0^infty mathrme^-s^2/4 mathrme^-z mathrmds \
&= 2mathrme^z^2 intlimits_0^infty mathrme^-(s+2z)^2/4 mathrmds \
&= 4mathrme^z^2 intlimits_0^infty mathrme^-y^2 mathrmdy \
&= 2sqrtpimathrme^z^2 mathrmerfc(z)
endalign
We completed the square in the exponent and used the substitution $y=z+s/2$.

Putting the pieces together yields our desired result
beginalign
frac2pi mathrme^-z^2 intlimits_0^infty fracmathrme^-z^2x^2x^2 + 1 mathrmdx &=
fraczpi mathrme^-z^2 intlimits_-infty^infty fracmathrme^-t^2z^2 + t^2 mathrmdt \
&= fraczpi mathrme^-z^2 frac1sqrt2 frac1z sqrtfracpi2 2sqrtpi mathrme^z^2 mathrmerfc(z) \
&= mathrmerfc(z)
endalign

With the substitution $x=fractz$, the integral on the left becomes

$$I=frac2pi z e^z^2int_0^+inftyfrace^-t^21+fract^2z^2,dt = frac1pi z e^z^2int_-infty^+inftyfrace^-t^21+fract^2z^2,dt $$
and we may switch to Fourier transforms. Since
$$mathcalF(e^-t^2) = frac1sqrt2e^-s^2/4,qquad mathcalFleft(frac11+fract^2z^2right)=zsqrtfracpi2 e^-z$$
$I$ boils down to an integral of the form $int_0^+inftyexpleft(-(s-xi)^2right),ds$ that is straightforward to convert in a expression involving the (complementary) error function.

As an alternative, you may use differentiation under the integral sign to prove that both sides of your equation fulfill the same differential equation with the same initial constraints, then invoke the uniqueness part of the Cauchy-Lipschitz theorem:
$$ fracddz LHS = -frac2piint_0^+infty2z e^-z^2 (x^2+1),dx,qquad fracddzRHS = -frac2sqrtpie^-z^2.$$
We have $fracddz(LHS-RHS)=0$, and $(LHS-RHS)(0)=1$.

An interesting consequence is the following (tight) approximation for the $texterfc$ function:

$$texterfc(z)=frac2e^-z^2piint_0^+inftyfrace^-z^2 x^2x^2+1,dxleq frac2e^-z^2piint_0^+inftyfracdx(x^2+1)(x^2 z^2+1)=frac1(1+z)e^z^2.$$

Assuming $z>0$,

$$ beginalignint_0^infty frace^-z^2x^21+x^2 , dx &= int_0^inftye^-z^2x^2 int_0^inftye^-t(1+x^2) , dt , dx \ &= int_0^infty e^-t int_0^inftye^-(z^2+t)x^2 , dx , dt tag1\ &= fracsqrtpi2int_0^infty frace^-tsqrtz^2+t , dt tag2\ &= fracsqrtpi2 , e^z^2int_z^2^inftyfrace^-usqrtu , du \ &= sqrtpi , e^z^2 int_z^infty e^-w^2 , dw \ &= fracpi2 , e^z^2operatornameerfc(z) endalign$$

$(1)$ Tonelli's theorem

$(2)$ $int_0^infty e^-ax^2 , dx = fracsqrtpi2 frac1sqrta$ for $a>0$

$(3)$ Let $u = z^2+t$.

$(4)$ Let $w=sqrtu$.

begineqnarray
&&intlimits_0^infty frace^-z^2 x^21+x^2 dx=\
&&intlimits_0^infty frace^-frac12 (sqrt2z)^2 x^21+x^2 dx=\
&&2 pi T(sqrt2 z, infty) e^frac12 (sqrt2z)^2\
&&2 pi intlimits_sqrt2 z^infty frace^-1/2 xi^2sqrt2 pi frac12 underbraceerf(fracinfty cdot xisqrt2)_1 dxi e^frac12 (sqrt2z)^2=\
&&fracpi2 erfc(z) e^z^2
endeqnarray
where $T(h,a)$ is the Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function .