Integration of natural log problem

kill00
Messages
8
Reaction score
0

Homework Statement


what is the integral of ln(e2x-1) dx

2. The attempt at a solution
I got
u= ln(e2x-1)
du = 2e2x-1 \ e2x-1
du = 2 dx

1/2(integral of u du)

1/2 ln (e2x-1 ) +c

=
1/4 [ln (e2x-1 )]2 + c

Could you please let me know if this is correct? I don't know how to integrate ln

thanks again
 
Last edited:
Physics news on Phys.org
Post deleted.
 
kill00 said:

Homework Statement


what is the integral of ln(e2x-1) dx

2. The attempt at a solution
I got
u= ln(e2x-1)
du = 2e2x-1 \ e2x-1
du = 2 dx

1/2(integral of u du)

1/2 ln (e2x-1 ) +c

=
1/4 [ln (e2x-1 )]2 + c

Could you please let me know if this is correct? I don't know how to integrate ln

thanks again

It might make your life a whole lot easier if you use the property that ln(ab) = b lna, and ln e = 1.
 
WOW, I seriously need to get some sleep if I couldn't think of that. I was studying for a physics test all night yesterday.

Thanks a lot gb7nash really appreciate it.
 
No problemo
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top