intergral

[expscv]

having problem with intergrate :

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


thx

[expscv]

btw how to use maths input thing?

[matt grime]

trig and latex, in that order.

[expscv]

huh? i m not sure wats exactly.

[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.

[matt grime]

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

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

[HallsofIvy]

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.

[expscv]

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


great thx

[matt grime]

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