How do you evaluate the integral $int 1/xdx$ from [-3,-2]?

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Answer 1 · 2017-02-23T21:40:12

Note that $int1/xdx=lnabsx+C$ .

So, when we apply the bounds, we see that:

$int_(-3)^(-2)1/xdx=[lnabsx]_(-3)^(-2)$

Evaluating this now:

$=lnabs(-2)-lnabs(-3)=ln2-ln3=ln(2/3)$

How do you evaluate the integral int 1/xdxint 1/xdx from [-3,-2]?