Why are you comparing it to anything? [MATH]\int {\frac{1}{{{{\left( {x - 2} \right)}^3}}}dx}=\dfrac{-1}{2(x-2)^2} [/MATH].It's the integral from 0 to infinity of dx/(x-2)^3. I don't know to what function should I compare it to know if it's divergent or convergent.
Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.Why are you comparing it to anything? [MATH]\int {\frac{1}{{{{\left( {x - 2} \right)}^3}}}dx}=\dfrac{-1}{2(x-2)^2} [/MATH].
Oops - I missed that! (So my post #3 is not correct - that is not all you have to do. You also have to investigate the existence of the integral at the vertical asymptote x=2).Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.
Are you saying "need" because you were instructed to do so? That is certainly not the only way, and probably not the best way.Because the result is convergent. However, I know it's divergent at x=2. So the only thing I need to know is if the integral from 2 to 3, for example, or from 0 to 2, is convergent or not. But I need to do it by comparison.