Question

$\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}\Rightarrow {\sin }^{-1}\left[\frac{A}{C}\right]=$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{6}$

$\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
$\Rightarrow x^2+2x+1=A(x^2+1)+x(Bx+C)$
Put $x=0\Rightarrow A=1$
Put $x=1\Rightarrow B+C=1$
Put $x=-1\Rightarrow B-C=-3$
On solving, we get $C=2$
Hence,${\sin }^{-1}(\frac{1}{2})=\frac{\pi }{6}$