$\int \frac{{\left(\cot \sqrt{{\tan }^{-1}x}\right)}^4\cos e{c}^2\sqrt{{\tan }^{-1}x}dx}{\left(1+x^2\right)\sqrt{{\tan }^{-1}x}}=$

$\frac{T={\displaystyle \int {\left[\cot \sqrt{{\tan }^{-1}x}\right]}^4\cos e{c}^2\sqrt{{\tan }^{-1}x}dx}}{\left(1+x^2\right)\sqrt{{\tan }^{-1}x}}$

$\text{ Put }\sqrt{{\tan }^{-1}x}=y$

$T=\int {\left(\cot y\right)}^4\cos e{c}^{2}ydx$

$\text{ Put }\cot y=z$

$\begin{array}{l}\cos e{c}^{2}ydy=-dz\\ T=\int z^4(-dz)\\ =\frac{-z^5}{5}+C\\ =\frac{-1}{5}(\cot y{)}^5+C\\ =\frac{-1}{5}{\cot }^5\sqrt{\tanh -1x}+C\end{array}$