Improper integral from 1 to infinity

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SUMMARY

The discussion centers on solving an improper integral from 1 to infinity, specifically focusing on the transformation of the expression $\frac{b}{b+1}$ into the form $1 - \frac{1}{b+1}$. The user initially derived the logarithmic expression $(\ln (b / (b+1)) - \ln (1 / (1+1)))$ but struggled to reach the boxed step. The correct manipulation of the fraction is confirmed as $\frac{b+1-1}{b+1} = 1 - \frac{1}{b+1}$, clarifying the steps needed to simplify the integral.

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with logarithmic properties
  • Basic algebraic manipulation skills
  • Knowledge of limits as they apply to infinity
NEXT STEPS
  • Study the properties of improper integrals in calculus
  • Review logarithmic identities and their applications
  • Practice algebraic manipulation of fractions
  • Explore limit concepts in calculus, particularly limits at infinity
USEFUL FOR

Students studying calculus, particularly those tackling improper integrals, as well as educators seeking to clarify concepts related to logarithmic transformations and algebraic simplifications.

yeny
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Hello everyone,

I am stuck on this homework problem. I got up to (ln (b / (b+1) - ln 1 / (1+1) ) but I'm not sure how to go to the red boxed step where they have (1 - 1 / (b+1) )

if anyone can figure it out Id really appreciate it.

thank you very much.
 

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$\frac{b}{b+1} = \frac{b+1-1}{b+1} = \frac{b+1}{b+1} + \frac{-1}{b+1} = 1 - \frac{1}{b+1}$
 
THANK YOU soo much for your help and time. Really appreciate it!
 

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