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

trig and latex, in that order.

 

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.

 

[tex] \int\frac{1}{x^2-2x-1}dx[/tex]

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

 

Getting back to the original problem, to integrate [itex] \int\frac{4}{x^2-2x-1}dx[/itex] 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 [itex]\int\frac{4}{u^2-2}du[/itex]. Now the denominator factors as
[itex]u^2-2= (u-\sqrt{2})(u+\sqrt{2})[/itex] and the integral can be done by partial fractions.

 

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