Hello,
you are supposed to know that:
$\displaystyle -\ln\left(\frac ab \right) = \ln(1)-\ln\left(\frac ab \right) = \ln\left(\frac{1}{\frac ab} \right) = \ln\left( \frac ba \right)$
Now consider your first equation:
1. Divide by $\displaystyle C_1$
2. Subtract the ln-value:
$\displaystyle -\ln\left(\frac{1+a}{1-a}\right)=C_2$ Use the property I mentioned above and you'll get
$\displaystyle \ln\left(\frac{1-a}{1+a}\right)=C_2$