integral involving natural log: integral of ln*sqrt(2x-1)

[Opus89]

Im working on an integral involving natural log and I'm running out of things to try. I'm almost certain I need to use parts and that's what I've been trying. I was also thinking along the lines of substitution but I'm not sure that could work in this case.

The integral is: the integral of ln*sqrt(2x-1).

I tried letting u = ln*sqrt(2x-1), du = 1/(2x-1), dv= dx, and v = x. This results in x* ln*sqrt(2x-1) - the integral of x/(2x-1). I'm not even sure this is the right way to go but I'm at a loss for what else I could do. Any hints?

 

[Deleted member 4993]

Re: integral involving natural log

Im working on an integral involving natural log and I'm running out of things to try. I'm almost certain I need to use parts and that's what I've been trying. I was also thinking along the lines of substitution but I'm not sure that could work in this case. The integral is: the integral of ln*sqrt(2x-1). I tried letting u = ln*sqrt(2x-1), du = 1/(2x-1), dv= dx, and v = x. This results in x* ln*sqrt(2x-1) - the integral of x/(2x-1). I'm not even sure this is the right way to go but I'm at a loss for what else I could do. Any hints?

substitute:

\(\displaystyle u \, = \, \sqrt{2x-1}\)

\(\displaystyle du \, = \, \frac{1}{2}\frac{2 \, dx}{\sqrt{2x-1}}\) ..... edited


\(\displaystyle dx \, = \, u \, du\)..... edited

 

[Opus89]

Re: integral involving natural log

if u = sqrt(2x-1), wouldn't the derivative, du = 1/sqrt(2x-1)? Unless I'm understanding this out of context.

 

[soroban]

Re: integral involving natural log

Hello, Opus89!

\(\displaystyle I \;=\;\int \ln\left(\sqrt{2x-1}\right)\,dx\)

\(\displaystyle \text{First, note that: }\:\ln\left(\sqrt{2x-1}\right) \;=\;\ln(2x-1)^{\frac{1}{2}} \;=\;\tfrac{1}{2}\ln(2x-1)\)

\(\displaystyle \text{So we have: }\;I \;=\;\tfrac{1}{2}\int \ln(2x-1)\,dx\)


\(\displaystyle \text{By parts: }\;\begin{array}{ccccccc}u &=& \ln(2x-1) & & dv&=& dx \\ du &=&\frac{2}{2x-1}\,dx & & v &=& x \end{array}\)


\(\displaystyle \text{Then we have: }\;I \;=\;\tfrac{1}{2}\bigg[x\ln(2x-1) - \int\frac{2x}{2x-1}\,dx\bigg]\)

. . . . . . . . . . . . \(\displaystyle =\;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}\int\left(1 + \frac{1}{2x-1}\right)dx\)

. . . . . . . . . . . . \(\displaystyle =\;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}\left[x + \tfrac{1}{2}\ln(2x-1)\bigg] + C\)

. . . . . . . . . . . . \(\displaystyle = \;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}x - \tfrac{1}{4}\ln(2x-1) + C\)

. . . . . . . . . . . . \(\displaystyle = \;\tfrac{1}{4}(2x-1)\ln(2x-1) - \tfrac{1}{2}x + C\)

 

[Deleted member 4993]

Re: integral involving natural log

if u = sqrt(2x-1), wouldn't the derivative, du = dx/sqrt(2x-1) ? Unless I'm understanding this out of context. <<< You are correct - I have edited my response above

 

[Deleted member 4993]

Re: integral involving natural log

\(\displaystyle \int ln\sqrt{2x-1}dx\) ............by substitution

Let

\(\displaystyle u \, = \, \sqrt{2x-1}\)

\(\displaystyle dx \, = \, u \, du\)

then we have

\(\displaystyle \int u\cdot ln(u)\, du\) ............now by parts

\(\displaystyle = \, \frac{u^2}{2}\cdot ln(u) - \int \frac{u^2}{2}\cdot \frac{1}{u}\, du\)

You know the rest....

Duplicate Post

viewtopic.php?f=3&t=32783&p=127491#p127491

 

[Opus89]

Re: integral involving natural log

Hello, Opus89!

\(\displaystyle I \;=\;\int \ln\left(\sqrt{2x-1}\right)\,dx\)

\(\displaystyle \text{First, note that: }\:\ln\left(\sqrt{2x-1}\right) \;=\;\ln(2x-1)^{\frac{1}{2}} \;=\;\tfrac{1}{2}\ln(2x-1)\)

\(\displaystyle \text{So we have: }\;I \;=\;\tfrac{1}{2}\int \ln(2x-1)\,dx\)


\(\displaystyle \text{By parts: }\;\begin{array}{ccccccc}u &=& \ln(2x-1) & & dv&=& dx \\ du &=&\frac{2}{2x-1}\,dx & & v &=& x \end{array}\)


\(\displaystyle \text{Then we have: }\;I \;=\;\tfrac{1}{2}\bigg[x\ln(2x-1) - \int\frac{2x}{2x-1}\,dx\bigg]\)

. . . . . . . . . . . . \(\displaystyle =\;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}\int\left(1 + \frac{1}{2x-1}\right)dx\)

. . . . . . . . . . . . \(\displaystyle =\;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}\left[x + \tfrac{1}{2}\ln(2x-1)\bigg] + C\)

. . . . . . . . . . . . \(\displaystyle = \;\tfrac{1}{2}x\ln(2x-1) - \tfrac{1}{2}x - \tfrac{1}{4}\ln(2x-1) + C\)

. . . . . . . . . . . . \(\displaystyle = \;\tfrac{1}{4}(2x-1)\ln(2x-1) - \tfrac{1}{2}x + C\)

I follow you all the way until you turn the integral of 2x/(2x-2) into the integral of 1 + 1/(2x-1). I'm just not seeing how those two steps are connected. I can follow everything else but that. I tried dividing the 2x/(2x-1) out but unless I did something wrong, that didn't help.

 

[Deleted member 4993]

Check my reply to your (same) question at:

viewtopic.php?f=3&t=32783&p=127491#p127491

 

[Opus89]

Ok, I finally got the answer. I see now how everything works in the problem. Thanks for your time and patience. I really appreciate it.