Question

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

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

Answer 1

Answer: (A)

$\frac{\pi }{6}$

$\frac{(x+1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
Consider, $\frac{(x+1{)}^2}{x^3+x}$
$=\frac{x^2+2x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
$\frac{x^2+2x+1}{x(x^2+1)}=\frac{A(x^2+1)+x(Bx+C)}{x(x^2+1)}$
$\frac{x^2+2x+1}{x(x^2+1)}=\frac{(A+B)x^2+Cx+A}{x(x^2+1)}$
$\Rightarrow x^2+2x+1=(A+B)x^2+Cx+A$

by comparing the coefficients we get

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

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

Hence, option 'A' is correct.