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

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About 9 years ago I prepared a detailed evaluation of this integral, and a .pdf file of my write-up can be found at this 14 Oct. 2009 Math Forum sci.math post. – Dave L. Renfro Jun 19 '12 at 14:51
Related problem (definite integral): (but not duplicate!) – apnorton Jan 14 '15 at 4:32

19 Answers 19

up vote 18 down vote accepted

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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$$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$


Set $x-\frac1x=\sqrt2\tan\phi$


Set $x+\frac1x=\sqrt2\sec\psi$

Reference: Trigonometric substitution

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My hint:

$$\int \frac{1}{1+x^4}dx=\frac{1}{2}\left[\int \frac{1+x^2}{1+x^4}dx+\int\frac{1-x^2}{1+x^4}dx\right]=\frac{1}{2}\left[\int \frac{1}{\left(x-\frac{1}{x}\right)^2+2}d\left(x-\frac{1}{x}\right)+\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x-\frac{1}{x}\right)\right]=\frac{1}{2}\left[\frac{1}{\sqrt{2}}\arctan\frac{x^2-1}{\sqrt{2}x}+\frac{}{2\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right]+Constant$$

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You can factor $x^4+1 = (x^2 + \sqrt{2} x + 1) (x^2 -\sqrt{2}x + 1)$. This allows you to write the integrand as $\frac{a_1 x + b_1}{x^2 + \sqrt{2}x + 1} + \frac{a_2 x + b_2}{x^2- \sqrt{2}x + 1}$. You would then rewrite the denominator in the form of $(x-u)^2 + v$ and rewrite the numerator as $a_i (x-u) + w$, from which you can do a change of variable to integrate essentially $x/(x^2+1)$ and $1/(x^2+1)$.

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

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HINT : Use $$x^4+1=(x^2+1)^2-(\sqrt 2x)^2=(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)$$ to get $$\begin{align}\int\frac{dx}{1+x^4}&=\int\frac{\frac{1}{2\sqrt 2}x+\frac 12}{x^2+\sqrt 2x+1}dx+\int\frac{-\frac{1}{2\sqrt 2}x+\frac 12}{x^2-\sqrt 2x+1}dx\\&=\frac{1}{2\sqrt 2}\int\frac{x+\sqrt 2}{x^2+\sqrt 2x+1}dx-\frac{1}{2\sqrt 2}\int\frac{x-\sqrt 2}{x^2-\sqrt 2x+1}dx\\&=\frac{1}{\sqrt 2}\int\frac{x+\sqrt 2}{(\sqrt 2x+1)^2+1}dx-\frac{1}{\sqrt 2}\int\frac{x-\sqrt 2}{(\sqrt 2x-1)^2+1}dx.\end{align}$$ Here, set $\sqrt 2x+1=\tan\alpha$ for the first and set $\sqrt 2x-1=\tan\beta$ for the second.

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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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Expand $\frac{1}{1+x^{4}}$ into partial fractions. For this purpose you need to factorize the polynomial in the denominator. You can proceed by writing it as a product of four linear terms \begin{equation*} x^{4}+1=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4}) \end{equation*} where $x_{1},x_{2},x_{3},x_{4}$ are its complex roots. Since \begin{equation*} x^{4}+1=0\Leftrightarrow x^{4}=-1=\cos \pi +i\sin \pi \end{equation*} the four roots are

\begin{eqnarray*} x_{1} &=&\cos \left( \frac{\pi }{4}\right) +i\sin \left( \frac{\pi }{4} \right) =\frac{\sqrt{2}}{2}(1+i) \\ x_{2} &=&\cos \left( \frac{3\pi }{4}\right) +i\sin \left( \frac{3\pi }{4} \right) =\frac{\sqrt{2}}{2}(-1+i) \\ x_{3} &=&\overline{x}_{2} \\ x_{4} &=&\overline{x}_{1}. \end{eqnarray*} Rewrite $x^{4}+1$ as a product of quadratic terms, by grouping the factors $ (x-x_{1}),(x-x_{4})=(x-\overline{x}_{1})$ and $(x-x_{2}),(x-x_{3})=(x- \overline{x}_{2})$

\begin{eqnarray*} x^{4}+1 &=&\left[ (x-x_{1})(x-\overline{x}_{1})\right] \left[ (x-x_{2})(x- \overline{x}_{2})\right] \\ &=&\left( x^{2}-\sqrt{2}+1\right) \left( x^{2}+\sqrt{2}x+1\right) \end{eqnarray*}

Find the constants $A,B,C,D$ such that \begin{equation*} \frac{1}{\left( x^{2}-x\sqrt{2}+1\right) \left( x^{2}+x\sqrt{2}+1\right) }= \frac{A+Bx}{x^{2}-x\sqrt{2}+1}+\frac{C+Dx}{x^{2}+\sqrt{2}x+1}. \end{equation*}

To evaluate \begin{equation*} \int \frac{1}{x^{2}\mp \sqrt{2}x+1}dx \end{equation*} complete the square in the denominator (see this answer of mine) \begin{equation*} x^{2}\mp \sqrt{2}x+1=\left( x\mp \frac{\sqrt{2}}{2}\right) ^{2}+\left( \frac{\sqrt{ 2}}{2}\right) ^{2} \end{equation*} and make the substitutions \begin{equation*} x\mp \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}t. \end{equation*} To compute the remaining integrals rewrite them as \begin{eqnarray*} \int \frac{x}{x^{2}\mp \sqrt{2}x+1}dx &=&\frac{1}{2}\int \frac{2x\mp \sqrt{2}}{ x^{2}\mp \sqrt{2}x+1}dx\pm\frac{\sqrt{2}}{2}\int \frac{1}{x^{2}\mp \sqrt{2}x+1}dx. \end{eqnarray*}

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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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Partial Fractions, and yes: you'll need arctangents.

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Or factor over the complex numbers, and your partial fractions will give you (complex) logarithms. – GEdgar Dec 27 '13 at 21:24

It is posiible to use faktorization $1+x^4=(1-\sqrt{2}x+x^2)(1+\sqrt{2}x+x^2)$ and partial fractions.

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Use $$1+t^4 = (1 + \sqrt{2} t + t^2 ) \times (1 - \sqrt{2} t + t^2 )$$ and decompose in partial fractions. You will arrive to much simpler antiderivatives.

I am sure that you can take from here.

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As in the question linked in Hans Lundmark's comment, you can factorise the denominator as $$x^4+1=(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)\ .$$ However you can make the resulting integral a bit less painful by getting all the surds outside the integral sign: first substitute $$u=x\sqrt2\ .$$ Then we have $$4(x^4+1)=u^4+4=(u^2+2u+2)(u^2-2u+2)$$ and the integral is evaluated by (relatively) simple partial fractions: $$\eqalign{I &=\int\frac{dx}{x^4+1}\cr &=2\sqrt2\int\frac{du}{(u^2+2u+2)(u^2-2u+2)}\cr &=\frac{1}{4\sqrt2} \int\Bigl(\frac{2u+4}{u^2+2u+2}-\frac{2u-4}{u^2-2u+2}\Bigr)du\cr &=\frac{1}{4\sqrt2} \bigl(\ln(u^2+2u+2)-\ln(u^2-2u+2)\cr &\qquad\qquad\qquad{}+2\tan^{-1}(u+1)+2\tan^{-1}(u-1)\bigr)+C\ .\cr}$$ You can now simplify the log and inverse tan terms (optional) and substitute back for $x$.

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


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:


After splitting the initial fraction we get:

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


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The steps are as follows:
1) Decompose $\frac{1}{1+x^{4}}$ using partial fractions (it can be factored using an identity of Sophie Germain)
2) You should have a linear function in each numerator and a quadratic in each denominator. Separate into the form $\frac{const}{quadratic}+\frac{const\cdot x}{quadratic}$
3) Complete the square on this quadratic.
4) To integrate the first form, make a simple substitution to transform the integrand into the form $\frac{1}{1+u^{2}}$, which is the derivative of $\tan^{-1}(x)$.
5) For the second, make another substitution to transform the integrand into the form $\frac{1}{1+v}$, which has antiderivative $\ln(1+v)$.

Be very careful with tiny algebraic slips, and keep track of your constants.

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HINT (for partial fractions) : $$ x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1). $$

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It is interesting that $\sqrt{2}$ appears suddenly. I did not manage to separate it into factors. – nesseril Dec 19 '12 at 12:30
$x^4+1$ has 4 complex roots (the 4th roots of $-1$) which come in pairs $w,\bar w$ and $z, \bar z$. Then one sees that $w+\bar w=\pm\sqrt{2}$ and $z+\bar z=\mp\sqrt 2$. – Andrea Mori Dec 19 '12 at 12:42
On a more elementary tune, consider the intermediate step $x^4+1=(x^2+1)^2-2x^2$. Here the appearence of $\sqrt2$ is immediately evident. – Andrea Mori Dec 19 '12 at 12:44

You can use $$x^4+1=(x^2+1)^2-2x^2=(x^2+1+\sqrt 2x)(x^2+1-\sqrt 2x).$$ Then, find $A,B,C,D$ such that $$\frac{1}{x^4+1}=\frac{Ax+B}{x^2+1+\sqrt 2x}+\frac{Cx+D}{x^2+1-\sqrt 2x}.$$ You'll find $$\frac{1}{x^4+1}=\frac{1}{2\sqrt 2}\left\{ \frac{x+\sqrt 2}{x^2+\sqrt2 x+1}+\frac{-x+\sqrt 2}{x^2-\sqrt 2x+1}\right\}.$$

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