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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Sign up to join this communityPlease, 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.
$$ \int \frac{dx}{x+\sqrt{x}} = \int \frac{1}{u^2+u} 2udu = 2\int \frac{1}{u+1}du $$ proceed..(using $u = \sqrt{x})$
$$ \int \frac {1}{x+\sqrt x}dx=2 \int \frac {\frac {1} {2 \sqrt x}} {\sqrt x +1}dx=2 \ln(\sqrt x +1)$$