Integration by Parts evaluation help

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SUMMARY

The discussion focuses on evaluating the integral of ln(2x + 1) using integration by parts. The user correctly identifies U as ln(2x + 1) and dv as 1dx, leading to the expression uv - ∫vdu. The user encounters difficulty at the integral ∫(x)(1/(2x + 1)) and receives a hint to simplify it using the identity 2x/(2x + 1) = 1 - 1/(2x + 1). This guidance is crucial for progressing towards the final solution.

PREREQUISITES
  • Understanding of integration techniques, specifically integration by parts
  • Familiarity with logarithmic functions and their properties
  • Basic differentiation skills, particularly with respect to composite functions
  • Knowledge of algebraic manipulation and simplification of fractions
NEXT STEPS
  • Practice additional integration by parts problems to reinforce the technique
  • Study the properties of logarithmic functions for better comprehension
  • Learn about integration techniques for rational functions, including partial fraction decomposition
  • Explore advanced integration methods such as substitution and trigonometric integrals
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators seeking to provide clearer explanations of integration by parts.

01010011
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Hi,
Can you tell me if I am on the right track with this problem. Thanks in advance.

Homework Statement


Evaluate the integral using integration by parts


Homework Equations


ln(2x + 1)dx


The Attempt at a Solution


ln(2x + 1)dx

= ln(2x + 1) * 1dx

Let U = ln(2x + 1)

therefore du = 1/(2x+1)

dv = 1dx

therefore v = x

Using the formula: \int udv = uv - \int vdu

= \int ln(2x + 1) * 1dx = ln (2x + 1)(x) - \int (x)(1/2x + 1)

If any of this is correct, I am now stuck at this point.
 
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01010011 said:
Let U = ln(2x + 1)

therefore du = 1/(2x+1)

Double check that differentiation :wink:

Other than that, it looks good so far. As a hint on how to continue, \frac{2x}{2x+1}=1-\frac{1}{2x+1}[/tex]
 

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