Question
Evalute : $\int \frac{1}{\sqrt{x}+x} d x$

Answer

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

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