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.
 A: Following @ClaudeLeibovici, we have 
$$I=y\log\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)-J$$
where 
$$\begin{align}
J&=\int\frac{y^2}{\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=-\int\frac{y^2\left(\frac{L}2-\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=-\frac{L}{2}\int\frac{y^2}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+\int\frac{y^2}{(y^2+z^2)}dy\\\\
&=-\frac{L}{2}\int\frac{1}{\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy+y-z\arctan(y/z) \\\\
&=K_1+K_2+y-z\arctan(y/z) 
\end{align}$$
where in $(1)$
$$K_1=-\frac{L}{2}\int\frac{1}{\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy$$
and 
$$K_2=\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy$$

We can easily evaluate $K_1$ by making the substitution $y=\sqrt{\left(\frac{L}{2}\right)^2+z^2}\tan t$.  Then, 
$$K_1=-\frac{L}{2}\log\left(y+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)$$
where the term $\sqrt{\left(\frac{L}{2}\right)^2+z^2}$ is an integration constant that we omitted.

We effect the same substitution for $K_2$.  Then, we have
$$\begin{align}
K_2&=\frac{L}{2}z^2\int\frac{1}{(y^2+z^2)\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}dy\\\\
&=\frac{L}{2}z^2\int \frac{\cos t}{\left(\frac{L}{2}\right)^2+z^2\sin^2t}dt\\\\
&=z\arctan\left(\frac{(L/2)\sin t}{z}\right)\\\\
&=z\arctan\left(\frac{(L/2)y}{z\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}\right)\\\\
\end{align}$$

Putting it all together reveals
$$\begin{align}
I&=y\log\left(\frac{L}2+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\\\\
&-y+z\arctan(y/z)\\\\
&+\frac{L}{2}\log\left(y+\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}\right)\\\\
&-z\arctan\left(\frac{(L/2)y}{z\sqrt{\left(\frac{L}{2}\right)^2+y^2+z^2}}\right)
\end{align}$$
A: 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 !
A: 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.
A: We can rework the constants to write the integral in the form
$$\int\ln\left(b\left(\sqrt{x^2+4}+a\right)\right)dx$$with $a<2$. The constant $b$ can be pulled out.
Set $x=t-\dfrac 1t$ to get
$$\int\ln\left(\sqrt{x^2+4}+a\right)dx=\int\ln\left(t+\frac1t+a\right)\left(1+\frac1{t^2}\right)dt.$$
You can decompose and transform the second term by inverting $t$:
$$\int\ln\left(t+\frac1t+a\right)\frac1{t^2}dt=-\int \ln\left(\frac1s+s+a\right)ds=\int\ln(s)\,ds-\int \ln\left(s^2+as+1\right)ds.$$
Now, by parts
$$\int\ln\left(t^2+at+1\right)dt=\frac12\left(2t+a\right)\ln\left(t^2+at+1\right)-\frac12\int\frac{4t^2+4at+1}{t^2+at+1}dt\\
=\frac12\left(2t+a\right)\ln\left(t^2+at+1\right)-2t+\frac32\int\frac{dt}{t^2+at+1},$$
and finally
$$\int\frac{dt}{t^2+at+1}=\frac4{4-a^2}\int\frac{dt}{\left(\dfrac{2t+a}{\sqrt{4-a^2}}\right)^2+1}=\frac1{\sqrt{4-a^2}}\arctan\left(\frac{2t+a}{\sqrt{4-a^2}}\right).$$
All together,
$$I(t)=\int\ln\left(t+\frac1t+a\right)dt=\frac12\left(2t+a\right)\ln\left(t^2+at+1\right)-t+\frac3{2\sqrt{4-a^2}}\arctan\left(\frac{2t+a}{\sqrt{4-a^2}}\right)-t\ln(t),$$
and $$\int\ln\left(b\left(\sqrt{x^2+4}+a\right)\right)dx=I\left(-\frac x2+\sqrt{\frac{x^2}4+1}\right)-I\left(-\frac x2-\sqrt{\frac{x^2}4+1}\right)+\ln(b)x.$$
