If int frac{sqrt{x}}{x(x+1)} d x = k tan^{-1} | vedclass

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(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = 	an^2 	heta$,which implies $dx = 2 	an 	heta \sec^2 	heta d	heta$. Then,$\sqrt{x}

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$(2, \sqrt{x})$

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(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = 	an^2 	heta$,which implies $dx = 2 	an 	heta \sec^2 	heta d	heta$. Then,$\sqrt{x} = 	an 	heta$. Substituting these into the integral: $I = \int \frac{	an 	heta}{	an^2 	heta (	an^2 	heta + 1)} \cdot (2 	an 	heta \sec^2 	heta) d	heta$. Since $	an^2 	heta + 1 = \sec^2 	heta$,the expression simplifies to: $I = \int \frac{	an 	heta}{	an^2 	heta \sec^2 	heta} \cdot (2 	an 	heta \sec^2 	heta) d	heta$. $I = \int \frac{2 	an^2 	heta \sec^2 	heta}{	an^2 	heta \sec^2 	heta} d	heta = \int 2 d	heta$. $I = 2	heta + C$. Since $x = 	an^2 	heta$,we have $	heta = 	an^{-1}(\sqrt{x})$. Therefore,$I = 2 	an^{-1}(\sqrt{x}) + C$. Comparing this with $k 	an^{-1} m$,we get $k = 2$ and $m = \sqrt{x}$. Thus,$(k, m) = (2, \sqrt{x})$.

If $\int \frac{\sqrt{x}}{x(x+1)} d x = k \tan^{-1} m$,then $(k, m)$ is

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