Question

If $f\text{'}\left(x\right)=x+\frac{1}{x}$ then the value of $f\left(x\right)$ is

• A. $x^2+\log x+c$
• B. $\frac{x^2}{2}+\log x+c$
• C. $\frac{x}{2}-\log x+c$
• D. None of these

Answer 1

The correct option is B

$\frac{x^2}{2}+\log x+c$

Explanation for the correct option:

Finding the value of $f\left(x\right)$:

Given that,

$f\text{'}\left(x\right)=x+\frac{1}{x}$

Integrate the above equation with respect to $x$,

$\begin{array}{rcl}f\left(x\right)& =& \int \left(x+\frac{1}{x}\right)dx\\ & =& \frac{x^2}{2}+\log x+c\left[\because \int x^{n}dx=\frac{x^{n+1}}{n+1}+c,\int \frac{1}{x}dx=\log x+c\right]\end{array}$

Hence, the correct option is B.