Improper Integrals with u sub and integration by parts ?

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SUMMARY

The discussion centers on evaluating the improper integral from 1 to infinity of (lnx)/(x) dx using u-substitution. The correct substitution is u = ln(x), leading to du = 1/x dx, which simplifies the integral to ∫u du = (1/2)(ln(x))^2 + C. To determine convergence, the integral must be evaluated as M approaches infinity, which is essential for concluding whether the integral diverges or converges.

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with u-substitution in integration
  • Knowledge of integration techniques, specifically integration by parts
  • Basic calculus concepts, including limits at infinity
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  • Study the evaluation of improper integrals, focusing on convergence tests
  • Learn more about u-substitution techniques in calculus
  • Explore integration by parts with practical examples
  • Investigate the behavior of logarithmic functions in integrals
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Students studying calculus, particularly those focusing on integration techniques and improper integrals, as well as educators seeking to clarify these concepts for their students.

Jay J
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Homework Statement



The integral from 1 to infinity of (lnx)/(x) dx

Homework Equations


U substitution and integration by parts


The Attempt at a Solution


Cant decide what to use as my "u" . . can anyone help with this part ?

-Jay J-
 
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this doesn't really require integration by parts, just a u-substitution:

\int \frac{ln(x)}{x}=\int u du = \frac{1}{2}u+ C= \frac{1}{2}(ln(x))^2+C

here you take u = ln(x) and differentiate to get, du = 1/x dx.
 
ok i understand the whole u substitution part but then i have to determine whether or not the integral is divergent or convergent and if its convergent i have 2 evaluate it . .. how would you do that ?

-Jay J-
 
Evaluate the integral between x=0 and x=M and let M approach infinity. What do you conclude?
 

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