Easy integration by parts help

MillerGenuine
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Homework Statement



integral of ln(2x +1)

Homework Equations


I know this is an easy problem but i can not seem to figure out what to substitute for my U and my dV. I was thinking on making my U= 2x+ 1. but then my problem is what would my dV be? ln U? lnx? the ln is throwing me off a bit. I am not sure what it is the ln of.


The Attempt at a Solution



U= 2x+1 dV= lnU (?)
dU/2= dx V= 1/U (?)
 
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When you want to integrate by parts, a good choice for u is that part which has a simple derivative, and for v that part which is easy to integrate. So you can choose ln(2x+1)=u and 1=dv/dx

ehild
 
ok perfect. got the answer. I have seen it done the way you just showed & i have also seen it done a different way. If we first substitute
p= 2x+1
dp/2 = dx
then once this is done we have 2 integral ln(p).
now from here we do integration by parts by choosing
u= ln(p) dV= dp
du= 1/p V= p

now is there any difference between the two? or any situations where one will work and the other wont?
 
Both methods have the same result, but substituting first 2x+1 by p is much simpler. Do not forget to plug in 2x+1 for p at the end.

ehild
 
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...

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