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$\frac{{{e^x}}}{{2x}}+ C$
$=\int {\frac{1}{x}(\ln x - \ln {e^x})\,dx}$
$= \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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$[A]$ $e^x-\int_0^x f(t) \sin t d t$ $[B]$ $x^9-f(x)$ $[C]$ $f(x)+\int_0^{\pi / 2} f(t) \sin t d t$
$[D]$ $x-\int_0^{\frac{\pi}{2}-x} f(t) \cos t d t$