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Can someone please help me find a useful substitution for the following integral: $$\int \frac{1}{\sqrt{x}(1+\sqrt{x})^2}dx$$ I tried letting $ u = \sqrt{x} $ But I couldn't proceed. Please help.

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Setting $1+\sqrt x=u$ gives you $$\frac{dx}{\sqrt x}=2du.$$ So, $$\int\frac{1}{(1+\sqrt x)^2}\cdot\color{red}{\frac{dx}{\sqrt x}}=\int\frac{1}{u^2}\cdot \color{red}{2du}.$$

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That's certainly a workable solution, as you can absorb the factor of $\sqrt{x}$ in the denominator when you write $du$, but you may as well take care of the constant in the parenthetical quantity and substitute

$$u = 1 + \sqrt{x},$$

so that $du = \frac{1}{2 \sqrt{x}}dx$, and your integral becomes

$$2 \int \frac{du}{u^2}.$$

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