Variable Change in Basic Integration

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SUMMARY

The discussion focuses on the change of variable method in basic integration, specifically using the integral \(\int \sqrt{\ln(x)}/x \, dx\). The variable substitution is defined as \(u = \ln(x)\) with the derivative \(du/dx = 1/x\). The integral simplifies to \(\int \sqrt{u} \, du\), leading to the result \(\frac{2}{3} u^{3/2} + C\), which translates back to \(\frac{2}{3} (\ln(x))^{3/2} + C\). The term \(1/x\) is not ignored; it is accounted for in the substitution process.

PREREQUISITES
  • Understanding of basic integration techniques
  • Familiarity with the change of variable method
  • Knowledge of derivatives and their role in integration
  • Ability to manipulate logarithmic functions
NEXT STEPS
  • Study the application of the substitution method in various integrals
  • Learn about integration techniques involving logarithmic functions
  • Explore advanced integration methods such as integration by parts
  • Practice solving integrals with different variable substitutions
USEFUL FOR

Students learning calculus, educators teaching integration techniques, and anyone seeking to improve their understanding of variable substitution in integrals.

Abraham
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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 because it's the derivative of ln(x)?

Thanks
 
Last edited:
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Abraham said:
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 because 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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