If int sin ^{-1}left(sqrt{frac{x}{1+x}}right) | vedclass

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(C) Let $x = 	an^2 	heta$,then $dx = 2 	an 	heta \sec^2 	heta d	heta$.Substituting this into the integral:$\int \sin^{-1}\left(\sqrt{\frac{	an^

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$(x+1, -\sqrt{x})$

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(C) Let $x = 	an^2 	heta$,then $dx = 2 	an 	heta \sec^2 	heta d	heta$.Substituting this into the integral:$\int \sin^{-1}\left(\sqrt{\frac{	an^2 	heta}{1+	an^2 	heta}}ight) (2 	an 	heta \sec^2 	heta) d	heta$$= \int \sin^{-1}(\sin 	heta) (2 	an 	heta \sec^2 	heta) d	heta = \int 	heta (2 	an 	heta \sec^2 	heta) d	heta$.Using integration by parts,let $u = 	heta$ and $dv = 2 	an 	heta \sec^2 	heta d	heta$. Then $du = d	heta$ and $v = 	an^2 	heta$.$= 	heta 	an^2 	heta - \int 	an^2 	heta d	heta$$= 	heta 	an^2 	heta - \int (\sec^2 	heta - 1) d	heta$$= 	heta 	an^2 	heta - (	an 	heta - 	heta) + C$$= 	heta (	an^2 	heta + 1) - 	an 	heta + C$$= (1+x) 	an^{-1}(\sqrt{x}) - \sqrt{x} + C$.Comparing this with $A(x) 	an^{-1}(\sqrt{x}) + B(x) + C$,we get $A(x) = x+1$ and $B(x) = -\sqrt{x}$.

If $\int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C$ where $C$ is a constant of integration,then the ordered pair $(A(x), B(x))$ can be

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