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Question

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

$\frac{5\pi}{6}$

0

$\frac{\pi}{6}$

$\frac{\pi}{2}$

Solution

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