Originally Posted by

**vince** obviously one can use integration by parts to attempt to solve any integral not just those with funky products in the intergrand so with that in mind i was trying to solve the integral,

$\displaystyle

\int\frac{1}{xln(x)}dx

$

using integration by parts after i had already noticed that direct substituion of u=ln(x) would do the trick.

But with the following choices this is what I'm getting:

Let f(x) = $\displaystyle \frac{dx}{x}$, g(x) = $\displaystyle \frac{1}{ln(x)}$

Then

$\displaystyle

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

= $\displaystyle 1 +\int\frac{1}{xln(x)}dx$

As you can see, this is nonsensical. Mr F says: No it's not. All it says is that the difference between the two integrals is a constant.

Would someone please tell me where I'm erring.

Thanks

The correct result is $\displaystyle

\int\frac{1}{xln(x)}dx = ln(ln(x)) + C $