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

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(A) Given the partial fraction decomposition: $\frac{(x + 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.Multiplying both sides by $x(x^2 +

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$\frac{\pi}{6}$

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(A) Given the partial fraction decomposition: $\frac{(x + 1)^2}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$.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:For $x^2$: $A + B = 1$.For $x$: $C = 2$.Constant term: $A = 1$.Substituting $A = 1$ into $A + B = 1$,we get $1 + B = 1$,so $B = 0$.Thus,$A = 1$ and $C = 2$.We need to find $\sin^{-1}\left(\frac{A}{C}\right) = \sin^{-1}\left(\frac{1}{2}\right)$.Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,the value is $\frac{\pi}{6}$.

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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