Indefinite Integration with Logarithms and Substitution

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SUMMARY

The discussion centers on solving the indefinite integral \(\int{\frac{1}{x \ln(x^3)}}dx\) using substitution and properties of logarithms. The participant initially attempted substitution with \(u=x^3\) but encountered difficulties due to the mismatch in terms. The key insight provided was recognizing that \(\ln(x^3) = 3\ln(x)\), leading to the solution of \(\frac{1}{3}\ln|\ln(x)|\). This highlights the importance of manipulating logarithmic expressions in integration.

PREREQUISITES
  • Understanding of indefinite integrals
  • Familiarity with substitution methods in calculus
  • Knowledge of logarithmic properties, specifically \(\ln(a^b) = b\ln(a)\)
  • Basic differentiation rules, particularly \(\frac{d}{dx} \ln(x) = \frac{1}{x}\)
NEXT STEPS
  • Practice solving more complex indefinite integrals using substitution
  • Explore advanced properties of logarithms in calculus
  • Learn techniques for integrating rational functions
  • Study integration by parts as a complementary method
USEFUL FOR

Students in calculus courses, educators teaching integration techniques, and anyone seeking to improve their skills in solving indefinite integrals involving logarithmic functions.

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Hi, I missed a few days of my calculus class. I've managed to figure out how to use substitution to solve an indefinite integral, and can apply the log properties to some extent. I just can't figure out this problem.

Homework Statement


Find the indefinite integral:
\int{\frac{1}{x ln(x^3)}}dx

2. The attempt at a solution
Well, since d/dx ln(x) is u'/u I know something is kinda wack with the bottom. I first tried to substitute with u=x^3, but then du is 3x^2 and there is only x on the bottom not x squared. I then thought it might be backwards, since the x on bottom is like x^-1 and it would have an integral with a natural log in it, but that wasn't really working out either.

I'm pretty stumped, I've attempted it several times.

Thanks for your help!
 
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Hint: ln(x^3)=3ln(x).
 
I hate it when that happens. so 1/3*ln|ln(x)|

Thanks grief!
 

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