$$\int(x^2-4)^{1/2} \, dx$$ using $$x=2\sec u, \,\,dx=2\sec u\tan u \,du$$
My answer: $$\frac{x(x^2-4)} 2 -2\ln\left|\frac{x+(x^2-4)^{1/2}}2 \right|$$ ...which I think is correct?
The book and calc says by the absolute value function applied to argument....gives the answer above without the 2.0 in the denominator in the 2nd "$|\ln|$" term...and I cannot see how they eliminated the 2.0 denominator
\begin{align} & \ln\left| \frac{\text{some function of } x} 2 \right| + \text{constant} \\[10pt] = {} & \ln\left|\text{the same function of }x\right| + \ln \frac 1 2 + \text{the same constant} \\[10pt] = {} & \ln\left|\text{the same function of }x\right| + \text{a different constant} \end{align}
But whether it's "the same constant" or "a different constant", either way the "constant" can be any number at all.