question_answer

\[\int{\frac{\sqrt{1+\sqrt{x}}}{x}}\,\,dx\] is equal to ________.

A)  \[4\sqrt{1+\sqrt{x}}+2\log \,\,\left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+C\]

B)  \[2\sqrt{1+\sqrt{x}}+4\log \,\,\left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+C\]

C)  \[3\sqrt{1+\sqrt{x}}+4\log \,\,\left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+C\]

D)  \[4\sqrt{1+\sqrt{x}}-2\log \,\,\left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+C\]

E)  None of these