Question

So Integral of ln x/x
What method should I use?
So far I used u'v+u'v and integrating by parts but it didnt work.

Answer 1

the answer is 1/2 (ln (x))^2 + c

Answer 2

okey I let u = 1/x probably why, okey Ill do it agian.

Answer 3

oh okey, what I forgot was to integrate again after doing u'v+u'v, thanx for the quick reply!

Answer 4

oh I didnt know that.

Answer 5

I did it here,
let u = t^2 --> du = 2t dt
dv = e^t dt--> v = e^t
the whole thing is
\[\int_0^2 t^2e^tdt=t^2e^t( from(0 to 2)) -\int_0^2 2te^tdt\]
=4e^2-\(2\int_0^2 te^tdt\)
the second part only:
\[2\int_0^2te^tdt\]

Answer 6

let u =t --> du =dt
dv=e^tdt--> v= e^t
\[2\int_0^2te^tdt=2[te^t(from 0to2)-\int_0^2e^tdt\]
=2[2e^2-e^2]=2e^2
replace to the whole thing above
=\(4e^2-2e^2=2e^2\)

Answer 7

@Xlegalize