Confusion about the integral $\int dT/(1-T^2)$

Question

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?

Answer 1

$$ \int \frac{dT}{1-T} = -\ln(1-T)+\text{constant}. $$ 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 $\ge0$.)

Answer 2

You need to pay attention to the constant of integration and don't ignore absolute value signs.

So you would have the same problem if you tried to do the same thing with this equaltiy $\frac{1}{1-x}=-\frac{1}{x-1}$

$\int \frac{1}{1-x}=-\int\frac{1}{x-1}$

$\log|1-x|=\log|x-1|+c_1$ Here you missed out the constant, also you are missing a minus sign

$|1-x|= c_2|x-1|$ In this step you missed out the absolute value signs.