# Integrate

• December 6th 2009, 04:39 PM
Morgan82
Integrate
Can someone start me out with this?

(lnx)^2 dx

I thought to make u = lnx, but then du = dx/x... I cant use it with that x, can I?

thanks!
• December 6th 2009, 04:42 PM
Jhevon
Quote:

Originally Posted by Morgan82
Can someone start me out with this?

(lnx)^2 dx

I thought to make u = lnx, but then du = dx/x... I cant use it with that x, can I?

thanks!

See here
• December 6th 2009, 04:45 PM
Prove It
Quote:

Originally Posted by Morgan82
Can someone start me out with this?

(lnx)^2 dx

I thought to make u = lnx, but then du = dx/x... I cant use it with that x, can I?

thanks!

Remember that $\ln^2{x} = \ln{x}\cdot\ln{x}$.

So use integration by parts.

You will need to make use of the fact that $\int{\ln{x}\,dx} = x\ln{x} - x$.