If frac{(x + 1)^2}{x^3 + x} = frac{A}{x} + fr | vedclass

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(A) Given $\frac{(x + 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Multiplying both sides by $x(x^2 + 1)$,we get $(x + 1)^2 = A(x^2 + 1)

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$\frac{\pi}{6}$

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(A) Given $\frac{(x + 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Multiplying both sides by $x(x^2 + 1)$,we get $(x + 1)^2 = A(x^2 + 1) + (Bx + C)x$.Expanding the left side: $x^2 + 2x + 1 = Ax^2 + A + Bx^2 + Cx$.Grouping terms: $x^2 + 2x + 1 = (A + B)x^2 + Cx + A$.Comparing coefficients:$A = 1$$C = 2$$A + B = 1 \implies 1 + B = 1 \implies B = 0$.Thus,$\sin^{-1}\left(\frac{A}{C}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.

If $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $\sin^{-1}\left(\frac{A}{C}\right) = $

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