Evaluating integral of 1/(x²-6x+8)

[radtad]

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

 

[dextercioby]

It's standard "arctan" type.To see it,try to write the denominator as a sum of squares...

Daniel.

 

[Jameson]

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.

$$\int\frac{1}{x^2-6x+8}dx = \int\frac{A}{x-4} + \frac{B}{x-2}dx$$

 

[dextercioby]

Sorry.


$$\int \frac{dx}{(x-3)^{2}-1}$$

and then the substituion $x-3=u$

which would give

$$\int \frac{du}{u^{2}-1}$$

which is typically "arctanh"...

Daniel.