Question

(intergral from 1 to infinity) [ln(x)]/[x^2] convergent or divergent?

Answer 1

i would say converge, but lets check it

Answer 2

I know it converges I just don't know how to show it does

Answer 3

find the anti derivative
think parts is what you need

Answer 4

i did parts and i got (-ln(x)/x) from one to infinity -1/x from one to infinity

Answer 5

put it over one denominator to make your life easier
then take the limit as x goes to infinity and see you get zero

Answer 6

so it would be (-ln(x)-1)/x

Answer 7

\[\lim_{x\to \infty}-\frac{\ln(x)+1}{x}=0\] for sure

Answer 8

why is the one not negative?

Answer 9

i pulled the minus sign out all the way
no difference

Answer 10

ok so we get negative infinity over infinity

Answer 11

ok now i suppose if you want to make your teacher happy you could appeal to l'hopital's rule, though it should be clear that \(\ln(x)\) grows much much more slowly than \(x\)

Answer 12

so it would be -1/x from 1 to infinity?

Answer 13

the limit will be zero
you still have to compute the integral, i.e. plug in 1

Answer 14

I don't understand. if we use l'hopital's rule to get -1/x from one to infinity i'm getting -1 as an answer

Answer 15

hold on i think i have confused you, let's take it step by step

Answer 16

oh nevermind I got it
-1/ infinity goes to zero then minus -1/1 haha brain fart

Answer 17

ok good got it

Answer 18

thank you so much

Answer 19

yw

Answer 20

is there a way I can thank you on here except for just saying thank you?