MCQ
If $\frac{d}{d x}(f(x))=\log x$, then $f(x)$ equals:
  • A
    $-\frac{1}{x}+C$
  • $x(\log x-1)+C$
  • C
    $x(\log x+x)+C$
  • D
    $\frac{1}{x}+C$

Answer

Correct option: B.
$x(\log x-1)+C$
We have, $\frac{d}{d x}(f(x))=\log x$
On integrating both sides, we get
$\int \frac{d}{d x}(f(x))=\int 1 \cdot \log x d x$
$\Rightarrow f(x)=\log x \int 1 \cdot d x-\int\left(\frac{d}{d x}(\log x) \int 1 \cdot d x\right) d x$
${[\text { Integrating by parts }]}$
$\Rightarrow f(x)=x \cdot \log x-\int \frac{1}{x} x x d x$
$\Rightarrow f(x)=x \log x-x+C$
$\Rightarrow f(x)=x(\log x-1)+C$

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