Need Help with Integral question

  • Thread starter radtad
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  • #1
radtad
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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
 
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  • #2
It's standard "arctan" type.To see it,try to write the denominator as a sum of squares...

Daniel.
 
  • #3
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]
 
  • #4
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.
 
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