How to integrate$ I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy $ I am stuck with the integration
$$
I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$$
I got this from the question from the book
"Field and wave electromagnetics, Cheng, 2nd, Problem 3-18.
I tried to solve this equation using method of integration by parts, but my equation got worse.
I know the answer by Wolfram Alpha, but I can't get how.

Considering $$I=\int\log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy$$ what I would first do is to get rid of the logarithm by a first integration by parts $$u=\log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)$$ $$du=\frac{y\, dy}{\sqrt{\frac{L^2}{4}+y^2+z^2}
   \left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)}$$ $v'=dy$, $v=y$ which makes $$I=y \log\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)- J$$  where $$J=\int\frac{y^2\, dy}{\sqrt{\frac{L^2}{4}+y^2+z^2}
   \left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)}$$ Now, an apparent change of variable could be $$\frac{L^2}{4}+y^2+z^2=t^2$$ $$y=\frac{1}{2} \sqrt{4 t^2-(L^2+4 z^2)}$$ $$dy=\frac{2 t}{\sqrt{4 t^2-(L^2+4 z^2)}}$$ which makes $$J=\int \frac{\sqrt{4 t^2-(L^2+4 z^2)}}{2t+L}\,dt$$ Again, $2t+L=w$, $t=\frac{w-L}{2}$, $dt=\frac{dw}{2}$ make $$J=\frac 12 \int\frac{\sqrt{w^2-2 L w-4 z^2}}{ w} dw$$ At this point, we find integrals which are given in the  Table of Integrals, Series, and Products by I.S. Gradshteyn and I.M. Ryzhik (in the $7^{th}$ edition, look at section $2.267$).
It is effectively quite tedious !
