$\int {(\frac{1 - tan x}{1 - tan x})}^{2} d x$

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Solution

$\int {(\frac{1 - t a n x}{1 - t a n x})}^{2} d x$
$= \int 1^{2} d x$
$= \int 1 d x$
$= \int d x$
$= x + c$

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