Integration of a natural log and polynomial

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Homework Help Overview

The discussion revolves around evaluating the indefinite integral of the natural logarithm of a polynomial expression, specifically ln(x² + 19x + 84), with the condition that x > 0. Participants explore various integration techniques and approaches to simplify the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts integration by parts, defining u and dv, and expresses concern over the correctness of their result. Another participant suggests factoring the quadratic to simplify the integral into two log terms. There is also a clarification regarding the nature of the integral, with a question about whether it is definite or indefinite.

Discussion Status

The discussion is active, with participants offering different methods and clarifications. One participant reports success with an alternative approach suggested by another, indicating a productive exchange of ideas. However, there is no explicit consensus on the best method yet.

Contextual Notes

There is a mention of the original poster's confusion regarding the type of integral being evaluated, which may affect the approach taken. Additionally, the problem involves a polynomial that may require specific techniques for integration.

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Homework Statement



Evaluate the integral when x > 0:

indefinite integral of ln(x2+19x+84)dx

Homework Equations



I know I need to use some form of integration by parts: integral of u*dv=uv-(integral of(du*v))

The Attempt at a Solution



I began by making u=ln(x2+19x+84) and dv=dx. Thus, (after u-substitution) du=(2x+19)/(x2+19x+84) and v=x.

After putting that in the formula, we get x*ln(x2+19x+84)-(integral of)((2x2+19x)/(x2+19x+84)). After simplifying that, I get:

x*ln(x2+19x+84)-((x2+19x+84)(4x+19)-(2x2+19x)(2x+19))/((x2+19x+84)2)

But according to the program I am using, that is the incorrect answer. Do you have any suggestions? Thanks.
 
Last edited:
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Why not just factor the quadratic, then split up the integral into two simpler log terms then use:

\int ln(u)du=u\ln(u)-u
 
Hmm, by infinite do you mean definite integral from 0 to infinity? If so, it's clearly divergent.
 
No, sorry, I meant the indefinite integral.
 
Thanks Jackmell. I tried that method and it worked. (A lot easier than the method I was using.)
 

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