1. ## Integration Question Help

I need to integrate this:

1/(x^4+4)

I have that it factors to (x^2+2x+2)(x^2-2x+2). I did the partial fraction decomposition correctly (I think), and I got this:

A=1/8
B=1/4
C=-1/8
D=1/8

I pulled 1/8 from the top of the integrations so I have to integrate:

(1/8)* the integral (x+2)/(x^2+2x+2) - (1/8)*the integral (x+2)/(x^2-2x+2).

That's where I'm stuck. Any help?

2. Try completing the square on the denominators. you should end up with some arctan in your solution.

3. Yeah, I tried. I don't know what to do from completing the square.

I get x^2+2x+1+1, which factors to (x+1)(x+1)+1.

On the other, I get x^2-2x+1+1, which factors to (x-1)(x-1)+1.

I don't know what to do with it.

4. $\displaystyle \frac{1}{8}\int\frac{2-x}{x^2-2x+1}dx=\frac{1}{8}\int\frac{2-x}{(x-1)^2+1}dx$

then let u= x-1

$\displaystyle \frac{1}{8}\int\frac{2-(u+1)}{u^2+1}du=\frac{1}{8}\int\frac{1}{u^2+1}du-\frac{1}{8}\int\frac{u}{u^2+1}du$

5. Shouldn't it be (x+2) on top and not (2-x)?

6. My partial fractions could be wrong but I came up with

$\displaystyle \frac{1}{x^4+4}=\frac{1}{8}\cdot\frac{-x+2}{x^2-2x+2}+\frac{1}{8}\cdot\frac{x+2}{x^2+2x+2}$

7. Originally Posted by thegame189
I need to integrate this:

1/(x^4+4)

I have that it factors to (x^2+2x+2)(x^2-2x+2). I did the partial fraction decomposition correctly (I think), and I got this:

A=1/8
B=1/4
C=-1/8
D=1/8

I pulled 1/8 from the top of the integrations so I have to integrate:

(1/8)* the integral (x+2)/(x^2+2x+2) - (1/8)*the integral (x+2)/(x^2-2x+2).

That's where I'm stuck. Any help?

You have the same answer as me

$\displaystyle \frac{cx+d}{x^2-2x+2}=\frac{1}{8}\cdot\frac{-x+2}{x^2-2x+2}$

8. Does this seem right?

(1/16)ln(x^2+2x+2) + (1/8)arctan(x+1) + (1/8)arctan(x-1) - (1/16)ln(x^2-2x+2) + C

Something else said it should be -(1/8)arctan(1-x) as the third part instead of +(1/8)arctan(x-1). Which is it?.

9. ^^ Those two forms are equivalent. The - is distributed inside the arctan in the second (this is allowed since arctan is an odd function).

edit: also, your solution is correct.

10. Yeah, that's what I was thinking. I just don't see why they'd write it with the negative. Oh well, thanks everyone. I was on the right track, I just got stuck after the square completing.