int {frac{{{e^{{{tan }^{ - 1}}sqrt x }}}}{{sq | vedclass

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(D) Let $t = 	an^{-1} \sqrt{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{1 + (\sqrt{x})^2} 	imes \frac{1}{2\s

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$2{e^{{{	an }^{ - 1}}\sqrt x }} + c$

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(D) Let $t = 	an^{-1} \sqrt{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{1}{1 + (\sqrt{x})^2} 	imes \frac{1}{2\sqrt{x}} = \frac{1}{1+x} 	imes \frac{1}{2\sqrt{x}}$.Thus,$dt = \frac{dx}{2\sqrt{x}(1+x)} = \frac{dx}{2(\sqrt{x} + x\sqrt{x})}$.This implies $\frac{dx}{\sqrt{x} + x\sqrt{x}} = 2 dt$.Substituting these into the integral:$\int e^t (2 dt) = 2 \int e^t dt = 2e^t + c$.Replacing $t$ with $	an^{-1} \sqrt{x}$,we get:$2e^{	an^{-1} \sqrt{x}} + c$.

$\int {\frac{{{e^{{{\tan }^{ - 1}}\sqrt x }}}}{{\sqrt x + x\sqrt x }}} dx = $

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