Question

The solution of integral: $\int {(\sqrt{x}+\frac{1}{\sqrt{x}})}^{2}dx$ is equal to

• A. $\frac{x^2}{2}+2x+\log \left|x\right|+C$
• B. $\frac{x^2}{2}+2+\log \left|x\right|+C$
• C. $\frac{x^2}{2}+x+\log \left|x\right|+C$
• D. $\frac{x^2}{2}+2x+2\log \left|x\right|+C$
• E. $\frac{x^2}{2}-2x+\log \left|x\right|+C$

Answer 1

The correct option is A $\frac{x^2}{2}+2x+\log \left|x\right|+C$

Let $I=\int {(\sqrt{x}+\frac{1}{\sqrt{x}})}^{2}dx$
$=\int (x+\frac{1}{x}+2)dx$
$=\frac{x^2}{2}+\log \left|x\right|+2x+C$