Question

Consider the integral 1 ln(x)/ (x)^2 dx from 1 to infinity.

Answer 1

I'm just having trouble computing the antiderivative. I have the lim t -->infinity of 1ln(x)/(x)^2 dx from 1 to t

Answer 2

What have you tried? Looks like a candidate for "Parts" to me.

Answer 3

is it 1/ln (x) or 1*ln(x)?

Answer 4

it's 1 ln(x)

Answer 5

divided by
X^2

Answer 6

Wondering what the 1 infront of the ln(x) is for... \[\large \int\limits \frac{ 1ln(x) }{ x^2 }dx\]
so you integrate by parts,
u = ln (x) du = 1/xdx
dv = 1/x^2 v = -1/x

Answer 7

integration by parts. express you integral in the form:
\[\int\limits_{}^{}u*dv\] proofs can show that it is equivalent to:
\[u*v - \int\limits_{}^{}v*du\]

[NOTE: u's and v's look similar on openstudy]

u = ln(x) ; dv = dx/x^2
du = dx/x ; v = -1/x

Answer 8

\[\large \int\limits \frac{ 1\ln(x) }{ x^2 }dx = uv-\int\limits vdu = -\frac{ \ln(x) }{ x }- \int\limits (-\frac{ 1 }{ x })(\frac{ dx }{x })\]

Answer 9

And from here im sure you can evaluate the integral :)

Answer 10

Okay, thank you both! :)

Answer 11

You can try using this as well, alternative method, similar one ;)
\[\int\limits \frac{lnx}{x^2}dx \\ \\ u=lnx~~~~~e^u=x \\ \\ du=\frac{1}{x}dx \\ \\ \int\limits u\frac{1}{e^u}du \\ \\ \text{IBP} \\ \\ J=u~~~~~dk=\frac{1}{e^u}du \\ \\ dJ=du~~~~~k=-e^{-u} \\ \\ -ue^{-u}+\int\limits \frac{1}{e^{-u}}du \\ \\ -\frac{u}{e^u}-\frac{1}{e^u} \\ \\ -\frac{lnx}{x}-\frac{1}{x}\]

Answer 12

IBP?

Answer 13

Oh IBP=integration by parts :)

Answer 14

Okay, i'll keep that in mind. Thanks! :)

Answer 15

A helpful Hint, you can also use this trick called ILATE = Inverse > logarithmic > algebraic > trigonometric > exponential when inputting for "u" . It comes in handy

Answer 16

I've never heard of that, I'll look it up on google though.

Answer 17

http://en.wikipedia.org/wiki/Integration_by_parts

Answer 18

I like ILATE better than LIATE, which was proposed by Herbert Kasube. it's much more efficient, IMO.