Integrate [1/(x^4 + 4)] dx

[Screwdriver]

Homework Statement

Well gentlemen, another year, another integral eh? Anyways,

$$\int \frac{1}{x^4+4}\,dx$$

I really want to do this without looking at Wolfram/Google.

Homework Equations

U-substitutions, parts, partial fractions

The Attempt at a Solution

Basically I tried to factor the denominator and then subtract something to make up the difference:

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

Then I also noticed that:

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

Now here's the iffy part; I know that

$$\frac{1}{(x^2 + 2)^2 - (2x)^2} \neq \frac{1}{(x^2 - 2x + 2)^2}$$

But I feel like If I can somehow combine those two things I can maybe decompose this into partial fractions. Yay or nay?

 

[╔(σ_σ)╝]

a^2 - b^2 = (a+b)(a-b)

 

[Screwdriver]

a^2 - b^2 = (a+b)(a-b)

Totally gotcha. How embarrassing...

[Edited for an algebraic error]

$$\int \frac{1}{(x^2 + 2)^2 - (2x)^2}\,dx = \int \frac{1}{(x^2 + 2x + 2)(x^2 - 2x + 2)}\,dx$$

$$\frac{1}{(x^2 + 2x + 2)(x^2 - 2x + 2)}= \frac{Ax+B}{x^2 + 2x + 2}+\frac{Cx+D}{x^2 - 2x + 2}$$

$$1=(Ax + B)(x^2 - 2x + 2) + (Cx + D)(x^2 + 2x + 2)$$

$$1=A(x^3 - 2x^2 + 2x)+B(x^2 - 2x + 2)+C(x^3 + 2x^2 + 2x)+D(x^2 + 2x + 2)$$

$$1=(A + C)x ^3 + (-2A + B + 2C + D)x^2 + (2A -2B +2C + 2D)x + (2B + 2D)$$

$$\end{matrix} \begin{matrix} 1& 0& 1& 0& |0\\ -2& 1& 2& 1& |0\\ 2& -2& 2& 2& |0\\ 0& 2& 0& 2& |1 \end{matrix}$$

Solving yielded A=1/8, B=1/4, C=-1/8 and D =1/4.

$$\int\frac{Ax+B}{x^2 + 2x + 2}+\frac{Cx+D}{x^2 - 2x + 2}\,dx = \frac{1}{8}\int\frac{x+2}{x^2 + 2x + 2}\,dx-\frac{1}{8}\int\frac{x-2}{x^2 - 2x + 2}\,dx$$

$$\frac{1}{8}\int\frac{x+2}{x^2 + 2x + 2}\,dx =\frac{1}{8}\int\frac{x}{x^2 + 2x + 2}\,dx + \frac{1}{4}\int\frac{1}{x^2 + 2x + 2}\,dx$$

and

$$-\frac{1}{8}\int\frac{x-2}{x^2 - 2x + 2}\,dx =-\frac{1}{8}\int\frac{x}{x^2 - 2x + 2}\,dx +\frac{1}{4}\int\frac{1}{x^2 - 2x + 2}\,dx$$

Another roadblock. How can I integrate those 4 integrals? Some of them look like lns and arctans...

 

[Apphysicist]

Looks like a good place to complete the square for at least one of them.

 

[vela]

You should check your calculations of A and C. I got A=1/8 and C=-1/8 in Mathematica.

 

[╔(σ_σ)╝]

Just out of curiousity, did you run into this integral while trying to integrate $$tan^{-1}(x)$$ ? I remember running into some messy algebra for that guy and your question looks similar.

 

[vela]

Another roadblock. How can I integrate those 4 integrals? Some of them look like lns and arctans...

Do something like

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

The first integral on the RHS you can do with a simple substitution.

 

[Screwdriver]

You should check your calculations of A and C. I got A=1/8 and C=-1/8 in Mathematica.

You're right. There's an extra D in the matrix; I'm going to edit that post to fix it.

Just out of curiousity, did you run into this integral while trying to integrate$$tan^{-1}(x)$$ ? I remember running into some messy algebra for that guy and your question looks similar.

No, I ran into it by itself :tongue2:

Using the suggestion put forth by Apphysicist:

$$\frac{1}{4}\int\frac{1}{x^2 + 2x + 2}\,dx = \frac{1}{4}\int\frac{1}{(x+1)^2+1}\,dx$$

$$u = x + 1$$
$$du = dx$$

$$\frac{1}{4}\int\frac{1}{(u)^2+1}\,du=\frac{1}{4}arctan(x +1)$$

And for the other one:

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

$$u = x - 1$$
$$du = dx$$

$$\frac{1}{4}\int\frac{1}{(u)^2+1}\,du=\frac{1}{4}arctan(x -1)$$

Using the suggestion put forth by vela, modified slightly:

$$\frac{1}{8}\int\frac{x}{x^2 + 2x + 2}\,dx =\frac{1}{16}\int\frac{2x+2}{x^2 + 2x + 2}\,dx - \frac{1}{8}\int\frac{1}{x^2 + 2x + 2}\,dx$$

I've already evaluated the right integral previously, it will be

$$-\frac{1}{8}arctan(x +1)$$

So for the left one:

$$\frac{1}{16}\int\frac{2x+2}{x^2 + 2x + 2}\,dx$$

$$u=x^2+2x+2$$
$$du=(2x+2)dx$$

$$\frac{1}{16}\int\frac{1}{u}\,dx$$
$$\frac{1}{16}ln(x^2+2x+2)$$

Last two:

$$-\frac{1}{8}\int\frac{x}{x^2 - 2x + 2}\,dx =-\frac{1}{16}\int\frac{2x-2}{x^2 - 2x + 2}\,dx -\frac{1}{8}\int\frac{1}{x^2 - 2x + 2}\,dx\\$$

I know that they're going to be

$$-\frac{1}{8}arctan(x -1)$$

and

$$-\frac{1}{16}ln(x^2-2x+2)$$

So FINALLY, the four arctans will combine leaving:

$$\int\frac{1}{x^4+4}\,dx = \frac{1}{8}arctan(x +1) +\frac{1}{8}arctan(x-1)+\frac{1}{16}ln(x^2+2x+2)-\frac{1}{16}ln(x^2-2x+2)+C$$

Man triumphs over machine.

 

[vela]

Now differentiate it and simplify to check your answer.

 

[SammyS]

From Screwdriver's 2^nd post.

$$\int\frac{Ax+B}{x^2 + 2x + 2}+\frac{Cx+D}{x^2 - 2x + 2}\,dx = \frac{1}{8}\int\frac{x+2}{x^2 + 2x + 2}\,dx-\frac{1}{8}\int\frac{x+2}{x^2 - 2x + 2}\,dx$$

$$\frac{1}{8}\int\frac{x+2}{x^2 + 2x + 2}\,dx =\frac{1}{8}\int\frac{x}{x^2 + 2x + 2}\,dx + 2\int\frac{1}{x^2 + 2x + 2}\,dx$$

and

$$-\frac{1}{8}\int\frac{x+2}{x^2 - 2x + 2}\,dx =-\frac{1}{8}\int\frac{x}{x^2 - 2x + 2}\,dx +2\int\frac{1}{x^2 - 2x + 2}\,dx$$

The next to last line should be:

$$\frac{1}{8}\int\frac{x+2}{x^2 + 2x + 2}\,dx =\frac{1}{8}\left(\int\frac{x}{x^2 + 2x + 2}\,dx + 2\int\frac{1}{x^2 + 2x + 2}\,dx\right)$$

The last line should be:

$$-\frac{1}{8}\int\frac{x+2}{x^2 - 2x + 2}\,dx =-\frac{1}{8}\left(\int\frac{x}{x^2 - 2x + 2}\,dx +2\int\frac{1}{x^2 - 2x + 2}\,dx\right)$$


I don't know if you carried this mistake through to the end.

 

[Screwdriver]

I don't know if you carried this mistake through to the end.

Turns out that not only did I do that, I also made about 10 other mistakes that took forever to find. I fixed and edited every post though - thanks for catching that mistake; my final answer was incorrect before

Now I'm positive that it's right

$$d/dx[\frac{1}{8}arctan(x+1) +\frac{1}{8}arctan(x-1)+\frac{1}{16}ln(x^2+2x+2)-\frac{1}{16}ln(x^2-2x+2)+C]$$

$$=\frac{1}{8(x^2+2x+2)}+\frac{1}{8(x^2-2x+2)}+\frac{x+1}{8(x^2+2x+2)}-\frac{x-1}{8(x^2-2x+2)}$$

$$=\frac{x+2}{8(x^2+2x+2)}-\frac{x-2}{8(x^2-2x+2)}$$

$$=\frac{1}{(x^2+2x+2)(x^2-2x+2)}$$

$$=\frac{1}{x^4+4}$$

Man triumphs over...self?