# Need Help with Integral question

1. Apr 10, 2005

How do you evaluate this integral:

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

i dont kno how to subsitute on this or anything. im completely stuck

2. Apr 10, 2005

### dextercioby

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

Daniel.

3. Apr 10, 2005

### 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.

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

4. Apr 10, 2005

### 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.