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.

This is incomplete.
But it's close.
We want
$I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$
If we look at
$\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}
=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{2}
$,
this is one of the roots
of the quadratic
$x^2-Lx-(y^2+z^2)
=0$.
Since the product of the roots is
$-(y^2+z^2)
$,
if we define
$J=\int\ln\left(\frac{L}{2}-\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$,
then
$I+J
=\int\ln(-(y^2+z^2))dy
$.
However,
the inner expression is negative,
so I'll define
$J=\int\ln\left(-\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$.
We now get
then
$I+J
=\int\ln(y^2+z^2)dy
$.
This is,
according to
Wolfram Alpha,
$y (log(y^2+z^2)-2)+2 z tan^{-1}(y/z)
$.
What we need now
is $I-J$.
The expression in the $\ln$
would be
$\begin{array}\\
\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}
&=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}\\
&=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}
\frac{L+\sqrt{L^2+4(y^2+z^2)}}{L+\sqrt{L^2+4(y^2+z^2)}}\\
&=\frac{L^2+L^2+4(y^2+z^2)+2L\sqrt{L^2+4(y^2+z^2)}}{4(y^2+z^2)}\\
&=\frac{L^2}{2(y^2+z^2)}+1+\frac{L\sqrt{L^2+4(y^2+z^2)}}{2(y^2+z^2)}\\
\end{array}
$
Again,
according to WA,
$\int \frac{\sqrt{x^2+a}}{x^2+b} dx 
= \frac{\sqrt{a-b} \tan^{-1}\left(\frac{x \sqrt{a-b}}{\sqrt{b} \sqrt{a+x^2}}\right)}{\sqrt{b}}+\ln(\sqrt{a+x^2}+x)
$
This would be enough to get
$I-J$,
and, from these,
$I$ amd $J$.
However, it's late
and I'm tired,
so I'll stop here.
