Question

\[\int\frac{1}{1 + \sqrt{x}} dx\]

Answer 1

\[\int \frac{dx}{1 + \sqrt{x}}\]
\[ = \int\frac{\sqrt{x} dx}{\sqrt{x} \left( 1 + \sqrt{x} \right)}\]
\[\text{Let 1} + \sqrt{x} = t\]
\[ \Rightarrow \sqrt{x} = t - 1\]

\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]

\[Now, \int\frac{\sqrt{x}}{\sqrt{x}\left( 1 + \sqrt{x} \right)}dx\]
\[ = 2\int\left( \frac{t - 1}{t} \right)dt\]
\[ = 2\int\left( 1 - \frac{1}{t} \right)dt\]
\[ = 2 \left( t - \text{log} \left| t \right| \right) + C\]
\[ = 2 \left( 1 + \sqrt{x} \right) - 2 \log \left| 1 + \sqrt{x} \right| + C\]
\[\text{Let} \text{ C }+ 2 = C'\]
\[ = 2\sqrt{x} - \text{2  log} \left( 1 + \sqrt{x} \right) + C'\]