[tex]\displaystyle\int\frac{\mathrm dx}{(1+x^2)\cot^{-1}x}[/tex]
Recall that [tex]\dfrac{\mathrm d}{\mathrm dt}\cot^{-1}t=-\dfrac1{1+t^2}[/tex]. So suppose we substitute [tex]y=\cot^{-1}x[/tex], so that [tex]\mathrm dy=-\dfrac{\mathrm dx}{1+x^2}[/tex]. Then the integral becomes
[tex]\displaystyle-\int\frac{\mathrm dy}y=-\ln|y|+C[/tex]
and replacing [tex]y[/tex] for [tex]x[/tex] gives the antiderivative
[tex]-\ln|\cot^{-1}x|+C=\ln\dfrac1{|\cot^{-1}x|}+C[/tex]
The proposed answer is not correct. Differentiating gives
[tex]\dfrac{\mathrm d}{\mathrm dx}\ln\dfrac1{\tan^{-1}x}=-\dfrac{\mathrm d}{\mathrm dx}\ln\tan^{-1}x=-\dfrac{\frac{\mathrm d}{\mathrm dx}\tan^{-1}x}{\tan^{-1}x}=-\dfrac1{(1+x^2)\tan^{-1}x}[/tex]
(Note the negative sign. We can also omit the absolute sign if we use a particular definition for [tex]\cot^{-1}x[/tex], but I don't think it's necessary to go into too much detail.)
