$\int \frac{\sqrt{x}}{x-1}dx $ I wished to solve $$\int \frac{\sqrt{x}}{x-1}dx \tag{1}$$ through hyperbolic trig substitution. My work is as follows:
Let $\sqrt{x}=\tanh (\theta)$. Then $(1)$ becomes $$\int -2\tanh^2(\theta)\ d\theta=-2\theta+2\tanh(\theta) +C.$$ Subbing back in yields $$-2\text{arctanh}\big(\sqrt{x}\big)+2\sqrt{x} +C. $$ Using $$\text{arctanh}(m)=\frac{1}{2}\ln\Big(\frac{1+m}{1-m} \Big) $$ I arrive at the answer $$2\sqrt{x}+\ln\Big(\frac{1-\sqrt{x}}{1+\sqrt{x}} \Big) +C. $$ However, the answer key and other integration methods suggest the answer $$ 2\sqrt{x}+\ln\Big(\frac{\sqrt{x}-1}{1+\sqrt{x}} \Big) +C. $$ Where did I go wrong? 

Strictly speaking, your answer is correct. If $x>0$, then $\ln(-x) = \ln(x)+\pi i$; these two are off by a constant. The difference between the two answers here is that what's in the $\ln$ is off by a factor of $-1$, and so the two functions are off by a constant. 
