How to show $\int_{\mathbb{R}}{t \choose x}^2{x \choose t}~dx = 1$

Let $${ a \choose b } = \frac{\Gamma(a+1)}{\Gamma(b+1)\Gamma(a-b+1)}$$ be the continuous extension of the binomial coefficient to non-integer arguments. I noticed this morning that $$\int_{\mathbb{R}}{t \choose x}^2{x \choose t}~dx = 1$$ For all real $t \geq0$. I tried to simplify the integrand according to the propeties on the MathWorld page for the gamma function, though I don't see where to apply even the reflection formula to:

$$\int_{\mathbb{R}}\frac{\Gamma(t+1)}{\Gamma(x+1)\,\Gamma(t-x+1)^2\,\Gamma(x-t+1)}~dx$$ To show that the above equals $1$. I couldn't find many similar questions on MSE about these sorts of integrals and am unfamiliar with how to approach them, so I bring this one here.

This is a direct application of the Ramanujan's Beta integral: \begin{align} \int_{-\infty}^{\infty}&\frac{\mathrm{d}x}{\Gamma\left(a+x\right)\Gamma\left(b+x\right)\Gamma\left(c-x\right)\Gamma\left(d-x\right)}=\\ &\hspace{3cm}\frac{\Gamma\left(a+b+c+d-3\right)}{\Gamma\left(a+c-1\right)\Gamma\left(a+d-1\right)\Gamma\left(b+c-1 \right)\Gamma\left(b+d-1\right)} \end{align} which is valid for $\Re\left( a+b+c+d \right)>3$.

Here, with $a=1,b=1-t,c=1+t,d=1+t$, we have $a+b+c+d=4+t>3$ if $t>-1$, and thus \begin{align} I(t)&=\int_{\mathbb{R}}\frac{\Gamma(1+t)}{\Gamma(x+1)\,\Gamma(1+t-x)^2\,\Gamma(x-t+1)}\,dx\\ &=\Gamma(1+t)\frac{\Gamma(1+t)}{\Gamma(1+t)\Gamma(1+t)\Gamma(1)\Gamma(1)}\\ &=1 \end{align} as expected. Then, this result should hold for $t>-1$.