<s>[INST] Confusion about the integral $\int dT/(1-T^2)$ From some reference on the internet we have the following real valued function and its derivative:
$$
M(T) = \frac{\sqrt{1-T^2}}{1+T}
\quad \Longrightarrow \quad \frac{dM}{dT} = - M/(1-T^2)
$$
The reverse of differentiation is integration:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
= \int \frac{dT}{T^2-1}\quad \Longrightarrow \\
\int\frac{dM}{M} = \frac{1}{2} \left[\, \int \frac{dT}{T-1} -\int \frac{dT}{T+1}\,\right]
\quad \Longrightarrow \\ \ln|M| = \frac{1}{2}\left[\,\ln|T-1|-\ln|T+1|\,\right] =
\ln\left(\sqrt{\left|\frac{T-1}{T+1}\right|}\right) \quad \Longrightarrow \\
M = \sqrt{\frac{1-T}{1+T}} = \frac{\sqrt{1-T^2}}{1+T}
$$
Here is where the confusion starts, because we also could have proceeded in the following way:
$$
\frac{dM}{dT} = - M/(1-T^2) \quad \Longrightarrow \quad \int\frac{dM}{M} = - \int \frac{dT}{1-T^2}
\quad \Longrightarrow \\
\int\frac{dM}{M} = - \frac{1}{2} \left[\, \int \frac{dT}{1-T} + \int \frac{dT}{1+T}\,\right]
\quad \Longrightarrow \\ \ln|M| = - \frac{1}{2}\left[\,\ln|1-T|+\ln|1+T|\,\right] =
\ln\left(\frac{1}{\sqrt{\left|1-T^2\right|}}\right) \quad \Longrightarrow \\
M = \frac{1}{\sqrt{1-T^2}}
$$
Which is clearly wrong. But .. where is the error?
 [/INST] $$
\int \frac{dT}{1-T} = -\ln(1-T)+\text{constant}.
$$
The initial minus sign comes from the chain rule.  You omitted it.
(No absolute value sign is needed since $\sqrt{1-T^2}$ exists only when $1-T$ and $1+T$ are both $\ge0$.)
</s>