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?