Integrate (1/x+2), where x= -6, x= -3
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I assume you're asking to find
$\displaystyle \int_{-6}^{-3}{\frac{1}{x + 2}\,dx}$
$\displaystyle = \left[\ln{|x + 2|}\right]_{-6}^{-3}$
$\displaystyle = \ln{|-6 + 2|} - \ln{|-3 + 2|}$
$\displaystyle = \ln{|-4|} - \ln{|-1|}$
$\displaystyle = \ln{4} - \ln{1}$
$\displaystyle = \ln{(2^2)} - 0$
$\displaystyle = 2\ln{2}$.
Let $\displaystyle u = x+2$, then $\displaystyle \dfrac{du}{dx} = 1$ $\displaystyle \Rightarrow {dx} = {du}$. When $\displaystyle x = -3$, $\displaystyle u = -1$; when $\displaystyle x = -6$, $\displaystyle u = -4$. So $\displaystyle \int_{-6}^{-3}\dfrac{1}{x+2}\;{dx} = \int_{-4}^{-1}\dfrac{1}{u}\;{du} = \bigg[\ln{|u|}\bigg]_{-4}^{-1} = \ln\left(1\right)-\ln\left(4\right) = \boxed{-\ln\left(4\right)} $, since $\displaystyle \ln\left(1\right) = 0$.