Assume that $X\sim\text{exp}(\beta),$ where $\beta$ itself is a random variable with $\sim\text{exp}(1).$ Determine the distribution of $\beta$ after we have observed $X=x.$
We have that $f_{X|\beta}(x|\beta)=\beta e^{-\beta x}$, $f_{\beta}(\beta)=e^{-\beta}$ and we are looking for $f_{\beta|X}(\beta|x).$ From Baye's we have that
$$f_{\beta|X}(\beta|x)=Cf_{X|\beta}(x|\beta)f_\beta(\beta)=C\beta e^{-\beta x}e^-{\beta}=C\beta e^{-(x+1)\beta}, \ \beta>0$$
and $C$ is a constant. This is a distribution with no famous name. I'd like to be done here but according to the book the answer is that
$$f_{\beta|X}(\beta|x)=(x+1)^2\beta e^{(x+1)\beta},$$
so $\beta\sim\Gamma(2,x+1)$. They come to this conclusion by integrating:
$$\int_0^{\infty}\beta e^{-(x+1)\beta} \ d\beta=\frac{1}{(x+1)^2}.$$
I have no idea what they are doing.
They are trying to find the value of the constant $C$. Since $f_{\beta \mid X}$ is a density it is the case that $$ \int_{0}^\infty C\beta e^{-(x+1)\beta}\, d\beta=1\implies C=\left[ \int_{0}^\infty \beta e^{-(x+1)\beta}\, d\beta \right]^{-1}=(x+1)^2. $$