Integration By Parts: Need help with a step

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SUMMARY

The integral of ln(2x + 1)dx is evaluated using integration by parts, resulting in the expression Xln(2x + 1) - ∫(2x/(2x + 1))dx. The confusion arises in simplifying the term 2x/(2x + 1) to the form [(2x + 1) - 1]/(2x + 1). This transformation is clarified as a simple arithmetic step where (2x + 1) - 1 simplifies to 2x, demonstrating that both forms are equivalent and useful for further integration.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts.
  • Familiarity with logarithmic functions and their properties.
  • Basic algebraic manipulation skills.
  • Knowledge of calculus concepts, particularly in evaluating integrals.
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  • Study the method of integration by parts in detail.
  • Practice simplifying rational expressions in integrals.
  • Explore the properties of logarithmic functions in calculus.
  • Review techniques for evaluating definite and indefinite integrals.
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Students preparing for Calculus II, educators teaching integration techniques, and anyone looking to strengthen their understanding of logarithmic integrals.

daviddee305
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Integration By Parts: Need help with a step...

Evaluate the integral:

\int ln(2x + 1)dx

I worked it out up until:

Xln(2x + 1) - \int 2x/(2x + 1) dx

Then the next step throws me off. I attached a scan from the solutions manual and circled the part that confused me. Could somebody explain how we went from "2x/(2x + 1)" to "[(2x + 1) - 1]/(2x + 1)"?

I'm sure it's a simple arithmetic step I'm overlooking, which is why I'm reviewing a few weeks before Calc. 2.


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Not much happens really, (2x+1)-1=2x+1-1=2x. As you can see both numerators are the same, the latter is just written in a more useful form.
 


thanks... it makes sense now...
 

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