Need Help with Integral question

  • Context: High School 
  • Thread starter Thread starter radtad
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Discussion Overview

The discussion revolves around evaluating the integral of the function \( \frac{1}{x^2-6x+8} \). Participants explore different methods for solving this integral, including substitution and partial fractions, while expressing varying opinions on the appropriate techniques to use.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses confusion about how to approach the integral and mentions being stuck.
  • Another participant suggests that the integral is a standard "arctan" type and recommends rewriting the denominator as a sum of squares.
  • A different participant counters that their use of Mathematica's integral calculator yielded a result involving the difference of two logarithms and proposes using partial fractions for the evaluation.
  • A later reply introduces a different integral, \( \int \frac{dx}{(x-3)^{2}-1} \), and suggests a substitution that leads to an integral typically associated with "arctanh".

Areas of Agreement / Disagreement

Participants present multiple competing views on how to evaluate the integral, with no consensus reached on the best approach or method.

Contextual Notes

There are unresolved assumptions regarding the methods suggested, and participants have not clarified the conditions under which their proposed techniques would be applicable.

radtad
Messages
19
Reaction score
0
How do you evaluate this integral:

integral(1/(x^2-6x+8)

i don't kno how to subsitute on this or anything. I am completely stuck
 
Physics news on Phys.org
It's standard "arctan" type.To see it,try to write the denominator as a sum of squares...

Daniel.
 
Standard Arctan? I did this on Mathematica's online integral calculator and got the difference of two logarithms. I would do this integral by partial fractions.

[tex]\int\frac{1}{x^2-6x+8}dx = \int\frac{A}{x-4} + \frac{B}{x-2}dx[/tex]
 
Sorry.:redface:


[tex]\int \frac{dx}{(x-3)^{2}-1}[/tex]

and then the substituion [itex]x-3=u[/itex]

which would give

[tex]\int \frac{du}{u^{2}-1}[/tex]

which is typically "arctanh"...

Daniel.
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 29 ·
Replies
29
Views
6K