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I am still confused with this integral: $\int \dfrac{1}{1+x^4}dx$

There one solution was given by a pupil that we write it as $\dfrac{x^2+1-(x^2-1)}{2(1+x^4)}$

Then split into two terms and proceed by putting $x+{1\over x}=t$ and completing the square.

My question is can we approach this question using other technique, most obviously partial fraction? The roots are complex, so how do we deal with it? Also please tell me a general procedure for such questions! Thank You!

jonsno
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  • You'll need to use techniques from complex analysis if you factor in terms of complex roots, specifically the residue theorem. – Matthew Cassell Sep 06 '16 at 16:15
  • Only partial fractions are required, as the integral is indefinite, so there is no need for evaluating residues. – poweierstrass Sep 06 '16 at 16:17
  • To use partial fractions with only real roots, we would have to note that $$ x^4 + 1 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1) $$ – Ben Grossmann Sep 06 '16 at 16:20
  • Related questions, maybe useful: http://math.stackexchange.com/questions/426152/evaluating-int-0-infty-fracdx1x4 http://math.stackexchange.com/questions/43457/how-can-i-compute-the-integral-int-0-infty-fracdt1t4?noredirect=1&lq=1 – Enrico M. Sep 06 '16 at 16:45
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    And this! http://math.stackexchange.com/questions/160157/integration-of-int-frac1x41-mathrm-dx – Enrico M. Sep 06 '16 at 16:46

2 Answers2

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$$x^4+1=(x^2+1)^2-2x^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$$

E.H.E
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2

For partial fractions factor into quadratics (this is the hard part): $$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$$

Do the standard decomposition method with undetermined cofficients:

$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} -\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1} \right)$$

Integrate: $$\int\frac{dx}{x^4+1}= \frac{1}{4\sqrt{2}}\ln \left(\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) +\frac{1}{2\sqrt{2}}((\arctan(\sqrt{2}x+1)+\arctan(\sqrt{2}x-1))$$

I think its quite beautiful.

Rene Schipperus
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