Integration of the reciprocal of the natural logarithm

[kudoushinichi88]

How do I start to evaluate this integral?

\int\frac{1}{\ln x}-\frac{1}{(\ln x)^2} dx

I tried subbing u=\ln x but I'm getting no where...

The answer is

\frac{x}{\ln x}+C[/itex]<br /> <br /> If I differentiate the answer, I get the integral easily, but the reverse... I&#039;m having trouble figuring out how do it.

 

[tiny-tim]

hi kudoushinichi88!

I tried subbing u=\ln x but I'm getting no where...

your substitution should have presented you with an easy integration by parts

but anyway you can do integration by parts on ∫ 1/(lnx)² dx

simply by first multiplying top and bottom by x: ∫ x/x(lnx)² dx