# Integration of $\int\frac{1}{x^{4}+1}dx$. [duplicate]

I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}dx$. Do I have to use trigonometric substitution?

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## marked as duplicate by Christoph, gerw, mau, Najib Idrissi, SabyasachiApr 4 at 8:48

You can use partial fractions. –  azarel Jun 19 '12 at 3:02
I tried partial fraction but i am stuck! –  Kns Jun 19 '12 at 3:03
Hasn't this been asked multiple times? –  The Chaz 2.0 Jun 19 '12 at 3:40
@The Chaz : probably, this looks like one of those integrals people get stuck on while beginning to learn how to integrate! –  Patrick Da Silva Jun 19 '12 at 4:02
This similar problem is listed in the related questions. –  Mark Hurd Jun 19 '12 at 7:05

I think you can do it this way.

\begin{align*} \int \frac{1}{x^4 +1} \ dx & = \frac{1}{2} \cdot \int\frac{2}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot \int\frac{(1-x^{2}) + (1+x^{2})}{1+x^{4}} \ dx \\\ &=\frac{1}{2} \cdot \int \frac{1-x^2}{1+x^{4}} \ dx + \frac{1}{2} \int \frac{1+x^{2}}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot -\int \frac{1-\frac{1}{x^2}}{\Bigl(x+\frac{1}{x})^{2} - 2} \ dx + \text{same trick} \end{align*}

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In fourth step you divide numerator and denominator by $x^{2}$? –  Kns Jun 19 '12 at 3:07
@Kns: Yes, I guess that was easy to find out. –  user33965 Jun 19 '12 at 3:08
+1 Nice trick there! –  DonAntonio Jun 19 '12 at 3:10

By partial fractions, $$\frac{1}{1+x^4} = \frac{1}{2\sqrt{2}}\left(\frac{x + \sqrt{2}}{x^2 + \sqrt{2}x + 1} - \frac{x - \sqrt{2}}{x^2 - \sqrt{2}x + 1}\right).$$ The rest is standard and not a great deal of fun. Complete the squares at the bottom and make the natural substitutions.

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the key is to show $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)=x^4+1$

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What Chandrasekhar wrote is a very nice trick. I'll offer you here a more "standard" one: $$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x-1)\Longrightarrow \frac{1}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x-1}$$and now do partial fractions and find the coefficients $\,A,B,C,D$

Added...or wait until someone else do it for you, of course.

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$$1=(Ax+B)(x^{2}-\sqrt{2}x-1)+(Cx+D)(x^{2}+\sqrt{2}x+1)$$ Now comparing the coefficients of $x^{3}$, we get $$A+C=0. \tag{1}$$ Comparing the constant coefficients, we get, $$-B+D=1.\tag{2}$$ Comparing the coefficients of $x^{2}$, we get $$B-\sqrt{2}A+D+\sqrt{2}C=0.\tag{3}$$ comparing the coefficients of $x$, we get $$-A-\sqrt{2}B+C+\sqrt{2}D=0. \tag{4}$$ Now by eq.$(2)$ we get, $$-A+C=-\sqrt{2}.\tag{5}$$ By, $(1)$ and $(5)$, we get $$A=\frac{1}{\sqrt{2}}, C=-\frac{1}{\sqrt{2}}$$ Similarly we can get $$D=\frac{3}{2}, B=\frac{1}{2}.$$ –  Kns Jun 19 '12 at 3:25

There are two (three) ways to go. One, assume

$$x^4+1=(x^2+ax+1)(x^2-ax+1)$$

You'll get that

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

Then $a=\sqrt 2$ (or the other, by symmetry)

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

The other ${x^2} = \tan \theta$, but it might get messy, unless you know how to use the Weierstrass substitution for example.

$$\int {\frac{{dx}}{{{x^4} + 1}}} = \int {\frac{{\left( {{{\tan }^2}\theta + 1} \right)d\theta }}{{{{\tan }^2}\theta + 1}}} \frac{1}{{2\sqrt {\tan \theta } }} = \int {\sqrt {\frac{{\cos\theta }}{{\sin\theta }}} \frac{{d\theta }}{2}}$$

$$\int {\sqrt {\frac{{\frac{{1 - {u^2}}}{{1 + {u^2}}}}}{{\frac{{2u}}{{1 + {u^2}}}}}} \frac{{du}}{{1 + {u^2}}}} = \int {\sqrt {\frac{{1 - {u^2}}}{{2u}}} \frac{{du}}{{1 + {u^2}}}}$$

However, Chandrasekar's is the best way to go, if you can figure it out.

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Does the last integral succesfully figures itself out or did you just not want to work it out and knew it works? –  Patrick Da Silva Jun 19 '12 at 4:00

Directly by Sophie Germain's Identity or:

$$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$$

After splitting the initial fraction we get:

$$\int \frac{1}{x^4 +1} \ dx = \int \frac{\frac{x}{2\sqrt2}+\frac{1}{2}}{x^2+\sqrt2x+1} \ dx+\int \frac{\frac{-x}{2\sqrt2}+\frac{1}{2}}{x^2-\sqrt2x+1} \ dx=$$ $$\frac{\sqrt2}{8} \int \frac{2x+2\sqrt2}{x^2+\sqrt2x+1} dx-\frac{\sqrt2}{8} \int \frac{2x-2\sqrt2}{x^2-\sqrt2x+1} dx=$$ $$\frac{\sqrt2}{8}\int \frac{2x+\sqrt2}{x^2+\sqrt2x+1} dx+\frac{1}{4} \int \frac{1}{x^2+\sqrt2x+1} dx-\frac{\sqrt2}{8}\int \frac{2x-\sqrt2}{x^2-\sqrt2x+1} dx+\frac{1}{4} \int \frac{1}{x^2-\sqrt2x+1} dx=$$ $$\frac{\sqrt2}{8}\left( \int \frac{2x+\sqrt2}{x^2+\sqrt2x+1} dx -\int \frac{2x-\sqrt2}{x^2-\sqrt2x+1} dx \right)+$$ $$\frac{\sqrt2}{4} \left( \int \frac{\sqrt2}{(\sqrt2x+1)^2+1} dx+\int \frac{\sqrt2}{(\sqrt2x-1)^2+1} dx \right)=$$ $$\frac{\sqrt2}{8} \left(\ln(x^2+\sqrt2x+1)-\ln(x^2-\sqrt2x+1) \right) +\frac{\sqrt2}{4} \left(\arctan(\sqrt2x+1)+ \arctan(\sqrt2x-1)\right)+C$$ $$=\frac{\sqrt2}{8} \ln\frac{(x^2+x\sqrt2+1)}{(x^2-x\sqrt2+1)}+\frac{\sqrt2}{4}\arctan\frac{x\sqrt2}{1-x^2}+C.$$

Q.E.D.

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