Question

integral from 0 to infinity of:
lnx/x^2

Answer 1

do you mean \[ \frac{ \ln x }{ x ^{2} }\]

Answer 2

HI!!

Answer 3

i would say parts for this one
to find the anti derivative i mean

Answer 4

tru \(u=\ln(x), dv =\frac{1}{x^2}\)

Answer 5

parts looks good

Answer 6

cause the next one will be a u-sub

Answer 7

\[\int\limits_{ }^{ } \frac{\ln(x)}{x^2}~dx\]\[u=\ln(x)~~~~~~~~~du=1/x~dx~~~~~~~~~~~1/x=1/e^u\]\[\int\limits_{ }^{ } \frac{u}{e^u}~du\]and then do parts.

Answer 8

that is my first look at it.

Answer 9

\[v=\int\limits_{ }^{ } e^{-u}~du=-e^{-u}\]\[u=u,~~~~>>~~du=1\]

Answer 10

\[\int\limits_{ }^{ } \frac{u}{e^u}~du=-\frac{u}{e^u}+\int\limits_{ }^{ }e^{-u}~du\]

Answer 11

then it should not be hard... after you have the answer in terms of u, sub the x back in

Answer 12

Why not by parts though? By parts is easy...

Answer 13

where did the e come from?

Answer 14

I am using by parts, but I make mit a bit simpler by u sub on the first step, no?

Answer 15

sateline, what do you mean?

Answer 16

No you are making it more complex I think.

Answer 17

it is easier to integrate e^(-u) and to differentiate the u.
and my sub was not that difficult.

Answer 18

you get
\[-\frac{\ln(x)}{x}+\int \frac{dx}{x^2}\] in one step

Answer 19

well... actually, yes. I change my mind.

Answer 20

it is easier to do by parts from the begin.

Answer 21

ln(x)/x is the thing that made me think of a u sub.... it is longer, but not a bad exercise.