# Having problem with intergrate : _ _/ 4 / (x^2-2x-1)

1. Mar 29, 2004

### expscv

having problem with intergrate :

... _
_/ 4 / (x^2-2x-1) dx

thx

Last edited: Mar 29, 2004
2. Mar 29, 2004

### expscv

btw how to use maths input thing?

3. Mar 29, 2004

### matt grime

trig and latex, in that order.

4. Mar 29, 2004

### expscv

huh? i m not sure wats exactly.

5. Mar 29, 2004

### matt grime

ok, to do the integral requires a trig subsitution, and to do the images, which is what I think you meant by "maths input thing" you need to use latex, which isnt' as odd as it might appear. there's a sticky thread somewhere explaining it, I beleieve if you go ot the general physics forum it's the first thread there. it takes a little memorizing but it's worth it.

6. Mar 29, 2004

### matt grime

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

click on the image and it should show you the source code for it

Last edited: Mar 29, 2004
7. Mar 29, 2004

### HallsofIvy

Staff Emeritus
Getting back to the original problem, to integrate $\int\frac{4}{x^2-2x-1}dx$ try completing the square in the denominator: x2- 2x- 1= x2- 2x+ 1- 2= (x-1)2-2. Make the change of variable u= x-1 and the integral becomes $\int\frac{4}{u^2-2}du$. Now the denominator factors as
$u^2-2= (u-\sqrt{2})(u+\sqrt{2})$ and the integral can be done by partial fractions.

8. Mar 29, 2004

### expscv

$$\int \frac{4}{x^2-2x-1} dx = \int \frac{4}{(x-1)^2-2}dx$$?

great thx

9. Mar 29, 2004

### matt grime

ah, apologies, it's hyperbolic trig, not trig.