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

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(D) Given,$\frac{(x+1)^{2}}{x^{3}+x} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1} \dots (i)$ Multiplying both sides by $x(x^{2}+1)$,we get: $(x+1)^{2} = A(x^{

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

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(D) Given,$\frac{(x+1)^{2}}{x^{3}+x} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1} \dots (i)$ Multiplying both sides by $x(x^{2}+1)$,we get: $(x+1)^{2} = A(x^{2}+1) + (Bx+C)x$ $x^{2} + 2x + 1 = Ax^{2} + A + Bx^{2} + Cx$ $x^{2} + 2x + 1 = (A+B)x^{2} + Cx + A$ Comparing the coefficients of $x^{2}$,$x$,and the constant term on both sides: $A+B = 1$ $C = 2$ $A = 1$ Substituting $A=1$ into $A+B=1$,we get $1+B=1$,so $B=0$. Now,we need to calculate $\sin^{-1} A + 	an^{-1} B + \sec^{-1} C$: $\sin^{-1}(1) + 	an^{-1}(0) + \sec^{-1}(2)$ $= \frac{\pi}{2} + 0 + \frac{\pi}{3}$ $= \frac{3\pi + 2\pi}{6} = \frac{5\pi}{6}$

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}$,then $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$ is equal to

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