\(\int\frac{1}{x+\sqrt{x}}dx\)=

\(log|1-\sqrt{x}|+c\) \(log|1+\sqrt{x}|+c\) \(2log|1-\sqrt{x}|+c\) \(2log|1+\sqrt{x}|+c\)

\(2log|1+\sqrt{x}|+c\)

\(x+\sqrt{x}=\sqrt{x}(1+\sqrt{x})\) Put \(1+\sqrt{x}=t\)