Practicing Success
The value of $\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 dx$ is: |
$\frac{x^2}{2}+\log |x|+C$, (where C is constant of integration) $\frac{x^2}{2}+\log |x|+2 x+C$, (where C is constant of integration) $\frac{3}{2}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+C$, (where C is constant of integration) $\frac{2}{3} x^{\frac{3}{2}}+2 \sqrt{x}+C$, (where C is constant of integration) |
$\frac{x^2}{2}+\log |x|+2 x+C$, (where C is constant of integration) |
$I=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 d x=\int(\sqrt{x})^2+\frac{1}{(\sqrt{x})^2}+2(\sqrt{x}) \times \frac{1}{(\sqrt{x})} d x$ $I=\int x+\frac{1}{x}+2 d x \Rightarrow I=\int x d x+\int \frac{1}{x} d x+\int 2 d x$ $\Rightarrow I=\frac{x^2}{2}+\log _e x+2 x+C$ |