I need help evaluating the following integral
$$\frac{ \sqrt 2 \sigma}{2 \epsilon_0} \int_0^R \frac{r \,dr}{ \sqrt{(z- \frac{ rh}{R} ) ^2 + r^2} }$$
This integral pertains to the Electric potential of an upside down hollow cone with height h and base radius R at an arbitrary point along the z axis. Apparently if I let $R = h$ the integral becomes:
$$\frac{ \sqrt 2 \sigma}{2 \epsilon_0} \int_0^R \frac{r \,dr}{ \sqrt{(z- r) ^2 + r^2} } = \frac{ \sqrt 2 \sigma}{2 \epsilon_0}[ \ln(1 + \sqrt2) - 1 ]$$
I am not sure what kind of substitution I should make if one at all. any suggestions would be nice. I tried wolfram alpha but It didn't work.
The integral you seek takes the general form
$$\int_0^R dr \: \frac{r}{\sqrt{r^2+a r+b}} $$
Rewrite this integral as follows:
$$\int_0^R dr \: \frac{r}{\sqrt{(r+a/2)^2+b-a^2/4}} $$
Now rewrite further as
$$\int_0^R dr \: \frac{r+a/2}{\sqrt{(r+a/2)^2+b-a^2/4}} - \int_0^R dr \: \frac{a/2}{\sqrt{(r+a/2)^2+b-a^2/4}}$$
You should be able to take it from here.