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Variable Change in Basic Integration

  • Thread starter Abraham
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  • #1
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Could someone please explain to me how to do basic integration using the change variable method?

I understand how to find "u" in the integrand, and then find du/dx, but I don't really know what's going on after that.

For example:

[tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x

u= ln(x), right?
and du/dx = 1/x

I know that the integral is equal to [tex]\sqrt{u}[/tex]= (2/3) ln(x)^(3/2)
But WHY? What happens to 1/x? Is it ignored becuase it's the derivative of ln(x)?

Thanks
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
5
Could someone please explain to me how to do basic integration using the change variable method?

I understand how to find "u" in the integrand, and then find du/dx, but I don't really know what's going on after that.

For example:

[tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x

u= ln(x), right?
and du/dx = 1/x

I know that the integral is equal to [tex]\sqrt{u}[/tex]= (2/3) ln(x)^(3/2)
But WHY? What happens to 1/x? Is it ignored becuase it's the derivative of ln(x)?

Thanks
[tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x
It should be written as
[tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x*dx
lnx = u
du = 1/x*dx
 

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