The value of $\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 dx$ is:
Answer & explanation
Correct answer: option 2
$I=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 d x=\int(\sqrt{x})^2+\frac{1}{(\sqrt{x})^2}+2(\sqrt{x}) \times \frac{1}{(\sqrt{x})} d x$
$I=\int x+\frac{1}{x}+2 d x \Rightarrow I=\int x d x+\int \frac{1}{x} d x+\int 2 d x$
$\Rightarrow I=\frac{x^2}{2}+\log _e x+2 x+C$