MCQ
Let $f (x) = \frac{1}{x}\,\ln \left( {\frac{x}{{{e^x}}}} \right)$ then its primitive $w.r.t. \,\,x$ is
  • A
    $\frac{1}{2} e^x - ln x + C$
  • B
    $\frac{1}{2} ln\, x - e^x + C$
  • $\frac{1}{2} ln^2x - x + C$
  • D
    $\frac{{{e^x}}}{{2x}}+ C$ 

Answer

Correct option: C.
$\frac{1}{2} ln^2x - x + C$
c
$\int {\frac{1}{x}\ln \frac{x}{{{e^x}}}\,dx}$

$=\int {\frac{1}{x}(\ln x - \ln {e^x})\,dx}$

$= \int {\frac{{\ln x - x}}{x}\,dx} $

$=\left[ {\int {\frac{1}{x}\ln \,x\,dx}  - \int {\frac{1}{x}x\,dx} } \right]$ (put $ln\, x = u$ ;$\frac{1}{x}dx = du$ )
$= \int {u\,dx}  - \int {1\,dx} $

$= \frac{1}{2} ln^2x - x + C$

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