$\int \frac{l o g (x + 2)}{(x + 2)^{2}} d x$ .

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Solution

Let $log (x + 2) = m$
$\frac{1}{x + 2} d x = d m$
$\int m e^{- m} d m$
$- m e^{- m} - e^{- m} + C$
$- \frac{log (x + 2)}{x + 2} - \frac{1}{x + 2} + C$

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