int frac{dx}{sqrt{x}+x} = | vedclass

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(D) To solve the integral $I = \int \frac{dx}{\sqrt{x} + x}$,we can simplify the denominator by factoring out $\sqrt{x}$.$I = \int \frac{dx}{\sqrt{x}(

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$2 \log (1 + \sqrt{x}) + c$

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(D) To solve the integral $I = \int \frac{dx}{\sqrt{x} + x}$,we can simplify the denominator by factoring out $\sqrt{x}$.$I = \int \frac{dx}{\sqrt{x}(1 + \sqrt{x})}$Let $u = 1 + \sqrt{x}$. Then,$du = \frac{1}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = 2 du$.Substituting these into the integral,we get:$I = \int \frac{2 du}{u} = 2 \int \frac{du}{u} = 2 \log |u| + c$.Substituting back $u = 1 + \sqrt{x}$,we obtain:$I = 2 \log (1 + \sqrt{x}) + c$.

$\int \frac{dx}{\sqrt{x}+x} = $

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