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

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(A) Given that $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$.Expanding the numerator on the left side: $\frac{x^{2}+2x+1}{x(x^{2}+1)}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given that $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$.Expanding the numerator on the left side: $\frac{x^{2}+2x+1}{x(x^{2}+1)} = \frac{x^{2}+1}{x(x^{2}+1)} + \frac{2x}{x(x^{2}+1)} = \frac{1}{x} + \frac{2}{x^{2}+1}$.Comparing this with $\frac{A}{x} + \frac{Bx+C}{x^{2}+1}$,we get $A=1$,$B=0$,and $C=2$.Now,we evaluate the expression $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C$.Note that $\cot^{-1}(\frac{1}{0})$ is $\cot^{-1}(\infty) = 0$.Substituting the values: $\operatorname{cosec}^{-1}(1) + \cot^{-1}(\infty) + \sec^{-1}(2) = \frac{\pi}{2} + 0 + \frac{\pi}{3} = \frac{5\pi}{6}$.

List-$I$	List-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

If $\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$,then $\operatorname{cosec}^{-1}\left(\frac{1}{A}\right)+\cot ^{-1}\left(\frac{1}{B}\right)+\sec ^{-1} C=$

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