If \( \frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1} \), then \( \operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C= \)

Given that \( \frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1} \rightarrow(1) \)
\[
\begin{array}{l}
\Rightarrow \frac{x^{2}+1+2 x}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{B x+C}{\left(x^{2}+1\right)} \\
\Rightarrow \frac{x^{2}+1}{\chi\left(x^{2}+1\right)}+\frac{2 x}{\chi\left(x^{2}+1\right)}=\frac{A}{\chi}+\frac{B x+C}{\left(x^{2}+1\right)} \\
\Rightarrow \frac{1}{\chi}+\frac{2}{\chi^{2}+1}=\frac{A}{\chi}+\frac{B x+C}{\left(x^{2}+1\right)}
\end{array}
\]
Comparing the coefficients, we have,
\[
A=1, B=0, C=2
\]
Now,
\[
\begin{array}{l}
\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C \\
\operatorname{cosec}^{-1}(1)+\cot ^{-1}\left(\frac{1}{0}\right)+\sec ^{-1}(2) \\
=\frac{\Pi}{2}+0+\frac{\Pi}{3}=\frac{5 \Pi}{6}
\end{array}
\]