another integration question (1 Viewer)

[echelon4]

here's 2 questions i can't seem to get the right answer to. (S=integral)

1. S 4dx/(x-1)^2(x+1)^2

2. S dx/(x^2-6x-6)

thanks in advanced.

 

[Rax]

Dont have time to do them right now but

You have to use partial fractions in the first one, and then when you integrate them it will involve logs

and the 2nd one you have to complete the square then possibly partial fractions aswell (and thus logs again)

Hope this helps

Cya

 

[LoneShadow]

1. 4/((x - 1)^2 (x + 1)^2) = -1/(x-1) + 1/(x-1)^2 + 1/(x+1) +1/(x+1)^2
So Integral = -2x/(x^2-1) + ln[(x+1)/(x-1)] + C

2. 1/(x^2-6x-6) = 1/[(x-3)^2-15]. Now let m = x-3. So dm = dx. Hence Integral[dx/(x^2-6x-6)] = Integral[dm/(m^2-(sqrt[15])^2)] = ln|[x-3-sqrt(15)]/[x-3+sqrt(15)]| + C


You should be able to do this questions. Don't just ask someone to do it for you the moment you get stuc on it.

 

[SeDaTeD]

Or, by inspection,
4/[(x-1)^2(x+1)^2]
= [1/(x-1)^2(x+1)^2]{(1+2x+x^2)+(1-2x+x^2)+(1+x-x^2-x^3)+(1-x-x^2+x^3)}
= [1/(x-1)^2(x+1)^2]{(1+x)^2 + (1-x)^2 - (x+1)^2(x-1) + (x-1)(x-1)^2}
= 1/(x-1)^2 + 1/(x+1)^2 - 1/(x-1) + 1/(x+1)

now integrate each separately.

 

[LoneShadow]

You do like confusing the kiddies, don't you? Just give him the most obvious method.

 

[SeDaTeD]

Partial fractions take too long. If there's a quicker way of splitting it up then go for it.

 

[LoneShadow]

Partial fractions doesnt take too long to figure using smart values for x. I haven't seen the method u have used here. Care to give a brief intro?

 

[SeDaTeD]

Called taking an educated/intuitive guess of what linear combination of those 4 'basis' polynomials yield 4. This was a lucky question anyway, it had a simple solution, it wouldn't work in general.

And I also prefer to use as few lines as possible.

 

[LoneShadow]

I see. Yeah fewer the no. lines the better.

 

[c0okies]

hey are you sure the 2nd question wasnt x^2-6x+8?

 

[SeDaTeD]

can be any quadratic.