Here follows a theorem from L. Evans PDE and a short part of its proof
Theorem
Assume $g\in C(\partial B(0,r))$, and define $u$ by
$$ u(x)=\frac{r^2-|x|^2}{n\alpha(n)}\int_{\partial B(0,r)}\frac{g(y)}{|x-y|^n}\,dy\qquad (x\in B^0(0,r)). $$ Then (i) $u\in C^\infty(B^0(0,r))$,
(ii) $\Delta u=0$ in $B^0(0,r)$, and
(iii) $\lim_{x\rightarrow x^0,x\in B^0(0,r)}u(x)=g(x^0)$ for each point $x^0\in\partial B(0,r)$
The proof of a similar theorem (for half-space) then states that after a direct calculation follows $$1=\int_{\partial B(0,r)}\frac{r^2-|x|^2}{n\alpha(n)r}\cdot\frac{1}{|x-y|^n}\,dS(y).$$
This is the part I don't get, I simply don't know how to show this. I can see we can take the first part out of the integral, but due to my lack of calculus skills, I don't know how to even begin with solving this integral.