Let $(X, Y)$ be a random variable in $\mathbb R^2$ with density $$f_{X, Y} (x, y) = \dfrac{\alpha (\alpha + 1)}{(1+x+y)^{2 + \alpha}} \mathbb 1_{[0, \infty)} (x) \mathbb 1_{[0, \infty)} (y)$$ for some parameter $\alpha \in \mathbb R_{>0}$.
Without doing computations, why do the two random variables $X$ and $Y$ have the same distribution?
Let's do the computations anyway.
$$f_X(x)=\int_{\mathbb{R}}f_{X,Y}(x,y)dy=\int_{(0,\infty)}\frac{\alpha(\alpha+1)}{(1+x+y)^{\alpha+2}}dy=\alpha(x+1)^{-(\alpha+1)},x \in [0,\infty)$$
$$f_Y(y)=\int_{\mathbb{R}}f_{X,Y}(x,y)dx=\int_{(0,\infty)}\frac{\alpha(\alpha+1)}{(1+x+y)^{\alpha+2}}dx=\alpha(y+1)^{-(\alpha+1)},y \in [0,\infty)$$
They are the same. But this was evident because $f_{X,Y}(x,y)=f_{X,Y}(y,x),\forall x,y \in \mathbb{R}^+\times\mathbb{R}^+$. Intuitively, it makes no real difference integrating the joint pdf with respect to $x$ or $y$.