Question
$\int_{}^{} {\frac{1}{x}\log x\;dx} $ =
$\log x = t $ रखने पर $ \Rightarrow \frac{1}{x}\,dx = dt$
$\therefore \,\,\,I\int_{}^{} {t\,dt} = \frac{{{t^2}}}{2} + c = \frac{{{{(\log x)}^2}}}{2} + c$.
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$x+(\cos \gamma) y+(\cos \beta) z=0$
$(\cos \gamma) x+y+(\cos \alpha) z=0$
$(\cos \beta) x+(\cos \alpha) y+z=0$