There is a ball of radius $a$ and a point $R$ inside this ball $(|R| <a)$ . Calculate the integral.
$$T\left(R\right)=\int_{r<a}^{ }\frac{d^{3}r}{\left|R-r\right|^{2}}$$
I tried to switch to spherical coordinates, but I don't understand what to do with the length of the vector in the denominator, what can you advise?
Without loss of generality, assume $R$ is on the $z$ axis. Then, in spherical coordinates, the integral becomes
$$I=\int_{r\lt a}\frac{ 1}{|R-r|^2}dr^3 = 2\pi \int_0^a\int_0^\pi \frac{r^2\sin\theta}{ R^2-2Rr\cos \theta +r^2}dr d\theta $$ Integrate over $\theta $ and then $r$ to obtain
\begin{align} I = \frac{2\pi}R\int_0^a r[\ln(r+R)-\ln|r-R|)]dr = \pi \left(\frac{a^2-R^2}R\ln\frac{a+R}{a-R}+2a\right) \end{align}
