The correct option is C x(log x 1)+c ∫ log x dx= ∫ log x 1 dx = log x∫ 1 dx ∫(^1/_x.√(1.dx))dx = log x(x) x+c = x(log x 1)+c ∵∫ ...

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Definition of Sets

∫logxdx=

Question

$\int log x d x =$

x (log x - x) + c

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x (log x) + c

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x (log x - 1) + c

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x (log x + x) + c

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Solution

The correct option is C $x (log x - 1) + c$
$\int l o g x d x = \int l o g x - 1 d x$
$= l o g x \int 1 - d x - \int (^{1} /_{x} . \sqrt{1. d x}) d x$
$= l o g x (x) - x + c$
$= x (l o g x - 1) + c$
$∵ \int l o g x d x = x (l o g x - 1) + c$

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