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)?
$\displaystyle
\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
\displaystyle
=-\frac{1}{2} \ \int x \ d \frac{1}{ln^2(x)}
$
You may use substitution (I usually don't use it)
$\displaystyle
\displaystyle
u=x ,\ \ v=\frac{1}{ln^2(x)}
$
We get
$\displaystyle
\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
\displaystyle
\frac{1}{ln^3(x)}> \frac{1}{x} \ .
$