The correct option is A x [ ( log x )^2 2 log x+2 ]. Integrating by parts, I=x ( log x )^2 2∫ x.log x.1/xdx =x ( log x )^2 2∫log x dx ...

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Bijective Function

∫ log x 2dx.

Question

$\int {(log x)}^{2} d x .$

x [{(log x)}^{2} - 2 log x + 2] .

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x [{(log x)}^{2} - 2 log x + 1] .

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x [{(log x)}^{2} - 2 log x - 2] .

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x [{(log x)}^{2} + 2 log x - 2] .

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Solution

The correct option is A $x [{(log x)}^{2} - 2 log x + 2] .$
Integrating by parts,
$I = x {(log x)}^{2} - 2 \int x . log x . \frac{1}{x} d x$
$= x {(log x)}^{2} - 2 \int log x d x$
$= x {(log x)}^{2} - 2 [x log x - x]$
$= x [{(log x)}^{2} - 2 log x + 2] .$

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