Possibly a Simple integration?

  • Thread starter metalmagik
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  • #1
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Main Question or Discussion Point

This integration might be very simple but I feel like I'm missing something here. I tried integration by parts but I ended up with a mess, x on top of x and the integral never came out without two x's as a product of each other.

[tex]\int[/tex][tex]\frac{ln(4x)}{2x}[/tex]

Any help is appreciated, I'm trying to learn Calculus myself.
 
Last edited:

Answers and Replies

  • #2
655
3
Hint: what is d/dx(ln(x))?
 
  • #3
107
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let [tex] 4x=y [/tex]

=>

[tex]\int\frac{ln 4x}{2x}dx = \frac{1}{2}\int\frac{lny}{y} dy[/tex]

integrate by parts :

[tex] lny = u , dv=\frac{dy}{y} [/tex]

=>

[tex] du=\frac{dy}{y} , v=lny [/tex]

=>

[tex]\int\frac{ln y}{y} dy = {(lny)}^2 - \int\frac{ln y}{y} dy [/tex]

=>

[tex]\int\frac{ln y}{y} dy = \frac{1}{2} {(lny)}^2 [/tex]

=>


[tex]\int\frac{ln 4x}{2x}dx = \frac{1}{4} {(ln4x)}^2 +C [/tex]
 
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  • #4
Gib Z
Homework Helper
3,346
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No, do NOT integrate by parts. Use maze's hint.
 
  • #5
107
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ok , maze's hint is smart , but i thought integrating by parts would be more informative as metalmagik said that he is learning !!
 
  • #6
131
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Whoa, I didn't even know anyone replied to this yet.

Well mmzaj's strategy looks like it works, but d/dx of ln(x) is 1/x...

if I use u substitution for this problem, I get u = ln(4x), du = 1/x dx

So, I have to put a 2 outside the integral to balance.

2[tex]\int[/tex]u du
2 [tex]\frac{u^2}{2}[/tex]

Then the 2's cancel out and I'm left with u^2 or ln(4x)^2. But this is not what mmzaj got from integration by parts.

Unless my u-substitution is off...

Why is there a problem?
 
  • #7
exk
119
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f(x)=ln(u) f'(x)=u'/u
 
  • #8
655
3
The 2 should be in the denominator when you pull it out, and then multiplying by the 1/2 from u^2/2, that gets the factor of 1/4.
 

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