\(\int\frac{1}{x+\sqrt{x}}dx\)=1\(log|1-\sqrt{x}|+c\)2\(log|1+\sqrt{x}|+c\)3\(2log|1-\sqrt{x}|+c\)4\(2log|1+\sqrt{x}|+c\)Answer & explanation+Correct answer: option 4\(x+\sqrt{x}=\sqrt{x}(1+\sqrt{x})\) Put \(1+\sqrt{x}=t\)
