Integration :[

Hey guys. This is just revision for me so I would appreciate it if you could tell me how to do the following. I really struggle with integrating ln :[

y=2ln(x-1)

y=2 over x^1/2(4lnx+3)

y=ln(3+2x^2)

Thanks :smile:

Reply 1

elldeegee

Original post by Robpattinsonxxx

use by parts where u= 2ln(x-1) and so can be differentiated.
and v=1, and so can be integrated.

Reply 2

f1mad

y=2ln(x-1)

Integration by parts: let u= ln(x-1) let dv/dx= 2.

y=2 over x^1/2(4lnx+3)

IBP again: pull the 2 out infront you're left with 1/x^1/2 * 1/4lnx+3

IBP again: -> y=ln(3+2x^2) is the same as y=ln(3+2x^2)*1

Let u= ln(3+2x^2) let dv/dx=1

Reply 3

Robpattinsonxxx

Thanks but I'm only on C3 at the moment so i don't know this integration by parts. Is there any other way?

That is C3 work.

Original post by Robpattinsonxxx

Thanks but I'm only on C3 at the moment so i don't know this integration by parts. Is there any other way?

This is c3. the method has been explained above.
yuo cannot integrate ln, which is why you need to do by parts, which is why such a question would come up in an exam to make sure you understand and remmeber the method.

EDIT: i told a lie, there is a way, but it isnt a way that would be done at alevel and so you wouldnt get the marks for it.