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Please, help me understand how to find

$$\int \frac{dx}{x+\sqrt{x}} = 2 \ln(\sqrt{x} + 1)$$

Is it done by some kind of substitution?

Note: by integrating the LHS, not differentiating RHS.

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    $\begingroup$ Try $\sqrt{x}=t$. $\endgroup$ – egreg Jul 7 '14 at 10:08
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    $\begingroup$ $$ \frac{\mathrm{d}}{\mathrm{d}x} 2\ln(\sqrt{x}+1) = 2\frac{\frac{1}{2 \sqrt{x}}}{\sqrt{x}+1}=\frac{1}{x+\sqrt{x}}$$ $\endgroup$ – jibounet Jul 7 '14 at 10:09
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$$ \int \frac{dx}{x+\sqrt{x}} = \int \frac{1}{u^2+u} 2udu = 2\int \frac{1}{u+1}du $$ proceed..(using $u = \sqrt{x})$

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$$ \int \frac {1}{x+\sqrt x}dx=2 \int \frac {\frac {1} {2 \sqrt x}} {\sqrt x +1}dx=2 \ln(\sqrt x +1)$$

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Try to use the substitution $\sqrt{x}=t$

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