- #1
- 69
- 0
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
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: