Mathematics

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


ANSWER

$$\dfrac {\pi}6$$


SOLUTION
$$\displaystyle \frac{(x+1)^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$$

Consider, $$\displaystyle \frac { (x+1)^{ 2 } }{ x^{ 3 }+x }$$

$$\displaystyle  =\frac { { x }^{ 2 }+2x+1 }{ x({ x }^{ 2 }+1) } =\frac { A }{ x } +\frac { Bx+C }{ x^{ 2 }+1 } $$

$$\displaystyle \frac { { x }^{ 2 }+2x+1 }{ x({ x }^{ 2 }+1) } =\frac { A(x^{ 2 }+1)+x(Bx+C) }{ x({ x }^{ 2 }+1) } $$

$$\displaystyle \frac { { x }^{ 2 }+2x+1 }{ x({ x }^{ 2 }+1) } =\frac { (A+B)x^{ 2 }+Cx+A }{ x({ x }^{ 2 }+1) } $$

$$\Rightarrow \displaystyle { x }^{ 2 }+2x+1=(A+B)x^{ 2 }+Cx+A$$

by comparing the coefficients we get

$$\Rightarrow A=1,C=2, A+B=1$$

$$\Rightarrow \sin^{-1}\left(\dfrac{A}{C}\right)= \sin^{-1}\left(\dfrac{1}{2}\right)=\displaystyle \dfrac{\pi}{6}$$

Hence, option 'A' is correct.
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