# Natural Logarithm Integration

• May 2nd 2010, 05:03 AM
creatively12
Natural Logarithm Integration
Integrate (1/x+2), where x= -6, x= -3

• May 2nd 2010, 05:11 AM
Prove It
Quote:

Originally Posted by creatively12
Integrate (1/x+2), where x= -6, x= -3

I assume you're asking to find

$\int_{-6}^{-3}{\frac{1}{x + 2}\,dx}$

$= \left[\ln{|x + 2|}\right]_{-6}^{-3}$

$= \ln{|-6 + 2|} - \ln{|-3 + 2|}$

$= \ln{|-4|} - \ln{|-1|}$

$= \ln{4} - \ln{1}$

$= \ln{(2^2)} - 0$

$= 2\ln{2}$.
• May 2nd 2010, 12:19 PM
creatively12
Thanks man, but the answer is -1n(4)
• May 2nd 2010, 03:52 PM
zzzoak
The 3 line is
$
\ln | -3+2 | - \ln | -6 +2 |
$
• May 2nd 2010, 04:06 PM
TheCoffeeMachine
Let $u = x+2$, then $\dfrac{du}{dx} = 1$ $\Rightarrow {dx} = {du}$. When $x = -3$, $u = -1$; when $x = -6$, $u = -4$. So $\int_{-6}^{-3}\dfrac{1}{x+2}\;{dx} = \int_{-4}^{-1}\dfrac{1}{u}\;{du} = \bigg[\ln{|u|}\bigg]_{-4}^{-1} = \ln\left(1\right)-\ln\left(4\right) = \boxed{-\ln\left(4\right)}$, since $\ln\left(1\right) = 0$.