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Integrate $\displaystyle \int \dfrac{1}{x^4+4}dx$.

I could try breaking this up into two quadratic trinomials, but that seems like it would be a lot of work. If that is the best way here how do I do it most efficiently?

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  • $\begingroup$ What's so hard about factoring into irreducible quadratics, then using partial fractions, then completing the square (twice), then substitution (twice), then... oh I get it. ;p $\endgroup$ – user137731 Dec 18 '15 at 15:12
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    $\begingroup$ "...but that seems like it would be a lot of work." Unfortunately, I'm not sure there's any way to get around that. This is definitely a bite-the-bullet kind of problem and I would attempt what user Bye_World has suggested above. $\endgroup$ – Xoque55 Dec 18 '15 at 15:19
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    $\begingroup$ You could always just use an integral table, though. $\endgroup$ – user137731 Dec 18 '15 at 15:19
  • $\begingroup$ WolframAlpha wolframalpha.com/input/?i=\int+\frac{1}{x^4+4} $\endgroup$ – StephenG Dec 18 '15 at 16:21
  • $\begingroup$ For a detailed solution, see the appropriate .pdf file at this 14 October 2009 sci.math post. See also these two math stackexchange questions: how to solve $\int\frac{1}{1+x^4}dx$ and Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$. $\endgroup$ – Dave L. Renfro Dec 18 '15 at 17:09
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I did it like this. You can Complete it Easily after this.

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Notice $$\frac{1}{x^4+4} = \frac{1}{(x^2+2)^2 - 4x^2} = \frac{1}{(x^2+2x+2)(x^2-2x+2)}$$ Since $$\begin{align} \frac{1}{x^2-2x+2} - \frac{1}{x^2+2x+2} &= \frac{4x}{(x^2+2x+2)(x^2-2x+2)}\\ \frac{1}{x^2-2x+2} + \frac{1}{x^2+2x+2} &= \frac{2x^2+4}{(x^2+2x+2)(x^2-2x+2)} \end{align}$$ We have

$$\frac{1}{(x^2+2x+2)(x^2-2x+2)} = \frac14\left[\frac{1-\frac{x}{2}}{x^2-2x+2} + \frac{1+\frac{x}{2}}{x^2+2x+2}\right]\\ = \frac18\left[\frac{x+2}{x^2+2x+2} - \frac{x-2}{x^2-2x+2}\right] = \frac18\left[\frac{(x+1)+1}{(x+1)^2+1} - \frac{(x-1)-1}{(x-1)^2+1}\right] $$ Up to an integration constant, this give us $$\begin{align} \int\frac{dx}{x^4+1} &= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right) + \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\ &= \frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right) + \frac18 \left[\tan^{-1}(x+1)+\tan^{-1}(x-1)\right]\\ &=\frac{1}{16}\log\left(\frac{(x+1)^2+1}{(x-1)^2+1}\right) + \frac18 \tan^{-1}\left(\frac{2x}{2-x^2}\right) \end{align} $$

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If you really want to avoid partial fractions, you can do it this way:

Expand in Taylor series (convergent for $|x| < \sqrt{2}$): $$ \dfrac{1}{4+x^4} = \dfrac{1/4}{1 + x^4/4} = \sum_{n=0}^\infty \dfrac{(-1)^n}{4^{n+1}} x^{4n} $$ Integrate term-by-term $$ \int \dfrac{1}{4+x^4} = \sum_{n=0}^\infty \dfrac{(-1)^n}{4^{n+1}} \dfrac{x^{4n+1}}{4n+1} = \dfrac{x}{16} \text{LerchPhi}\left(-\dfrac{x^4}{4},1,\dfrac{1}{4}\right)$$ where $$ \text{LerchPhi}(z,a,v) = \sum_{n=0}^\infty \dfrac{z^n}{(n+v)^a}$$ Of course you might want to convert that LerchPhi expression to something more elementary...

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HINT:

$$\int\frac{1}{x^4+4}\space\text{d}x=$$ $$\int\frac{1}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}\space\text{d}x=$$ $$\int\left(\frac{2-x}{8(x^2-2x+2)}+\frac{x+2}{8(x^2+2x+2)}\right)\space\text{d}x=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{x+2}{x^2+2x+2}\space\text{d}x=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\left(\frac{2x+2}{2(x^2+2x+2)}+\frac{1}{x^2+2x+2}\right)\space\text{d}x=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{x^2+2x+2}\space\text{d}x+\frac{1}{16}\int\frac{2x+2}{x^2+2x+2}\space\text{d}x=$$


For the integrand $\frac{2x+2}{x^2+2x+2}$, substitute $u=x^2+2x+2$ and $\text{d}u=(2x+2)\space\text{d}x$:


$$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{x^2+2x+2}\space\text{d}x+\frac{1}{16}\int\frac{1}{u}\space\text{d}u=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{x^2+2x+2}\space\text{d}x+\frac{\ln\left|u\right|}{16}=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{x^2+2x+2}\space\text{d}x+\frac{1}{16}\int\frac{1}{u}\space\text{d}u=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{x^2+2x+2}\space\text{d}x+\frac{\ln\left|x^2+2x+2\right|}{16}=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{(x+1)^2+1}\space\text{d}x+\frac{\ln\left|x^2+2x+2\right|}{16}=$$


For the integrand $\frac{1}{(x+1)^2+1}$, substitute $s=x+1$ and $\text{d}s=\text{d}x$:


$$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{1}{8}\int\frac{1}{s^2+1}\space\text{d}s+\frac{\ln\left|x^2+2x+2\right|}{16}=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{\arctan\left(s\right)}{8}+\frac{\ln\left|x^2+2x+2\right|}{16}=$$ $$\frac{1}{8}\int\frac{2-x}{x^2-2x+2}\space\text{d}x+\frac{\arctan\left(x+1\right)}{8}+\frac{\ln\left|x^2+2x+2\right|}{16}$$

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