Question
Evaluate the following functions : $\int \frac{1}{x+\sqrt{x}} \cdot d x$

Answer

$
\begin{aligned}
& \quad \mathrm{I}=\int \frac{1}{x+\sqrt{x}} \cdot d x \\
& =\int \frac{1}{\sqrt{x}(\sqrt{x}+1)} \cdot d x \\
& \text { put } \sqrt{x}+1=t \\
& \therefore \quad \frac{1}{2 \sqrt{x}} \cdot d x=1 \cdot d t \\
& \therefore \quad \frac{1}{\sqrt{x}} \cdot d x=2 \cdot d t \\
& =\int \frac{1}{t} \cdot 2 \cdot d t \\
& =2 \cdot \int \frac{1}{t} \cdot d t \\
& =2 \cdot \log (t)+c \\
& =2 \cdot \log (\sqrt{x}+1)+c
\end{aligned}
$

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