problem solving logarithmic equation and reaching an equivalence ok so i've had a problem trying to simplify the  $\ln\left[ \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right]$ and this is supposed to be equal to :  $\ln [ \sqrt{a^2+u^2} + u ]$
how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?
 A: I expect that you got this as the result of an (indefinite) integration, and $a$ is a constant. Let $a$ be positive. We are taking the ln of
$$\frac{1}{a}\left(\sqrt{u^2+a^2}+u\right).$$
Taking the ln, we get
$$\ln\left(\sqrt{a^2+u^2}+u\right)-\ln a.$$
But $\ln a$ is a constant, so can be absorbed into the constant of integration. 
In more detail, if
$$\ln\left(\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}\right)+C$$
is the answer to an indefinite integral problem, where $C$ is an arbitrary constant, then
$$\ln\left(\sqrt{a^2+u^2}+u\right)+D$$
is a correct answer to the same problem.
This sort of thing happens a lot, particularly with trigonometric functions. As a simple example, if $\sin^2 x+C$ is "the" answer to an indefinite integration problem, then so is $-\cos^2 x+C'$.
A: $\ln x$ is injective so, if $\ln x=\ln y\implies x=y$. So, the following should hold true:\begin{align}
\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}=\sqrt{a^2+u^2}+u\end{align}
However, simply plugging in $u=1, a=2$ gives us
\begin{align*}
&\sqrt{1+\frac{1}{4}}+\frac{1}{2}=\sqrt{4+1}+1\\
&\implies \sqrt{\frac{5}{4}}+\frac{1}{2}=\sqrt{5}+1\\
&\implies \sqrt{\frac{5}{4}}=\sqrt{5}-\frac{1}{2}\\
&\implies \frac{\sqrt{5}}{2}=\frac{2\sqrt{5}-1}{2}\\
&\implies \sqrt{5}=2\sqrt{5}-1
\end{align*}
which is false.
Hence they are not the same.
A: \begin{align}
& \ln\left( \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right) = \ln\left(\sqrt{\frac{a^2}{a^2}+\frac{u^2}{a^2}} + \frac u a\right) =  \ln\left(\sqrt{\frac{a^2+u^2}{a^2}} + \frac u a\right) \\[10pt]
= {} & \ln\left(\frac {\sqrt{a^2+u^2}} a  + \frac u a\right) = \ln\left(\frac{\sqrt{a^2+u^2}+u} a\right) = \ln\left(\sqrt{a^2+u^2} + u\right) - \ln a
\end{align}
If this is viewed as a function of $u$, then $a$ is constant, so it's
$$
\ln\left(\sqrt{a^2+u^2} + u\right) + \text{constant}
$$
