MCQ
The value of $\int \log x d x$ :
  • A
    $\log x-x+c$
  • B
    $1+\log x+c$
  • $x(\log x-1)+c$
  • D
    $x(\log x+1)+c$

Answer

Correct option: C.
$x(\log x-1)+c$
(C)
$\int_{\text {II }}^1 1 \cdot \log x \cdot d x=\log x \cdot x-\int \frac{1}{x} \times x d x$
$=x \log x-x=x(\log x-1)+c$

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