<s>[INST] 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 ?
 [/INST] \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
$$
\ln\left(\sqrt{a^2+u^2} + u\right) + \text{constant}
$$
</s>