$\int x (\frac{sec 2 x - 1}{sec 2 x + 1}) d x$ .

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Solution

$\int x (\frac{1 - cos 2 x}{1 + cos 2 x}) d x$
$\int x {tan}^{2} x d x$
$x sec x - \int sec x d x$
$x sec x - ln | sec x + tan x | + C$

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