Integral question

**Kuma** · March 28th 2011, 06:14 AM

Hi

I want to integrate the following:

1/(ln(x)^3) from 2 to infinity.

Am i allowed to do:

(ln(x)^-3)
-3 ln(x)
pull out -3 and integrate ln (x)?

**tonio** · March 28th 2011, 06:49 AM

Originally Posted by Kuma

Hi

I want to integrate the following:

1/(ln(x)^3) from 2 to infinity.

Am i allowed to do:

(ln(x)^-3)
-3 ln(x)
pull out -3 and integrate ln (x)?

Of course you're allowed, but be sure the power is on x and NOT on the whole ln(x)...

Tonio

**FernandoRevilla** · March 28th 2011, 07:05 AM

Originally Posted by Kuma

1/(ln(x)^3) from 2 to infinity.

Kuma: even if you don't use $\LaTeX$ code, you can avoid the ambiguous notation writing:

1 / [ ln (x^3) ] or 1 / [ ( ln x )^3 ]

which are different functions .

**Kuma** · March 28th 2011, 07:57 AM

Ah its the entire [ln(x)]^3.
What do i do in this case? Substitution?

**zzzoak** · March 28th 2011, 09:00 AM

$ \displaystyle \int \frac{dx}{ln^3(x)}=\int \frac{x \ dx}{xln^3(x)}=\int \frac{x \ d(lnx)}{ln^3(x)}=\int x \ \frac{(-1)}{2} \ d \frac{1}{ln^2(x)}= $

$ \displaystyle =-\frac{1}{2} \ \int x \ d \frac{1}{ln^2(x)} $

You may use substitution (I usually don't use it)

$ \displaystyle u=x ,\ \ v=\frac{1}{ln^2(x)} $

We get

$ \displaystyle =-\frac{1}{2} \ [ \ x \ \frac{1}{ln^2(x)}-\int \frac{1}{ln^2(x)} \ dx \ ] $

You may continue in the same way.

Also we may notice x>>1

$ \displaystyle \frac{1}{ln^3(x)}> \frac{1}{x} \ . $

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