Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?  
 A: $$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$
$$\int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{\left(x-\frac1x\right)^2+2}dx$$
Set $x-\frac1x=\sqrt2\tan\phi$
$$\int\frac{1-x^2}{1+x^4}dx=-\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$
Set $x+\frac1x=\sqrt2\sec\psi$
Reference: Trigonometric substitution 
A: HINT :
$$x^4+1
 =(x^2+1)^2-2x^2
 =(x^2+\sqrt 2x+1)(x^2-\sqrt2x+1)$$
A: My hint:
$$\int \frac{1}{1+x^4}dx=\frac{1}{2}\left[\int \frac{1+x^2}{1+x^4}dx+\int\frac{1-x^2}{1+x^4}dx\right]=\frac{1}{2}\left[\int \frac{1}{\left(x-\frac{1}{x}\right)^2+2}d\left(x-\frac{1}{x}\right)+\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x-\frac{1}{x}\right)\right]=\frac{1}{2}\left[\frac{1}{\sqrt{2}}\arctan\frac{x^2-1}{\sqrt{2}x}+\frac{}{2\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right]+Constant$$
A: 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 $1/(x^2+1)$. 
A: 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.
A: 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. 
A: 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*}
A: 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 $x$.
A: the key is to show $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)=x^4+1$
A: 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.
A: It is posiible to use  faktorization
$1+x^4=(1-\sqrt{2}x+x^2)(1+\sqrt{2}x+x^2)$
and partial fractions.
A: Hint
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.
A: 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*}
A: 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.
A: 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.
A: Hints:
$$x^4+1=(x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)$$
Partial Fractions, and yes: you'll need arctangents.
A: 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.
A: HINT (for partial fractions) :
$$
x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1).
$$
A: 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
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt 2}\left\{ \frac{x+\sqrt 2}{x^2+\sqrt2 x+1}+\frac{-x+\sqrt 2}{x^2-\sqrt 2x+1}\right\}.$$
A: Let $$I=\int\frac{dx}{x^4+1}$$
Enforce the substitution $x:=\frac{1}{y}\implies dx=-\frac{dy}{y^2}$ so that $$I=-\int\frac{dy}{y^2\left(\frac{1}{y^4}+1\right)}=-\int\frac{dy}{y^2+\frac{1}{y^2}}=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\tag1$$
Then observe that $$y^2+\frac{1}{y^2}=\left(y-\frac{1}{y}\right)^2+2=\left(y+\frac{1}{y}\right)^2-2\tag2$$
Now splitting $I$ at the end of $(1)$ into two integrals $$I=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}+1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy-\frac{1}{2}\int\frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}}dy\tag3$$ 
Observe $$\frac{d}{dy}\left(y+\frac{1}{y}\right)=1-\frac{1}{y^2}$$ and $$\frac{d}{dy}\left(y-\frac{1}{y}\right)=1+\frac{1}{y^2}$$ The latter integrals in $(3)$ now become by combining the last identities & $(2)$:$$I=-\frac{1}{2}\int\frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2-2} dy-\frac{1}{2}\int\frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2+2} dy=-\frac{1}{2}\left(\int\frac{dt}{t^2-2}+\int\frac{du}{u^2+2}\right)$$
upon the substitutions $t=y+\frac{1}{y}$ and $u=y-\frac{1}{y}$.
$$I=-\frac{1}{2}\left(\int\frac{dt}{-2\left(1-\left(\frac{t}{\sqrt{2}}\right)^2\right)}+\int\frac{du}{2\left(\left(\frac{u}{\sqrt{2}}\right)^2+1\right)}\right)=-\frac{1}{4}\left(\int\frac{du}{\left(\frac{u}{\sqrt{2}}\right)^2+1}-\int\frac{dt}{1-\left(\frac{t}{\sqrt{2}}\right)^2}\right)$$ $$=-\frac{\sqrt{2}}{4}\left(\int\frac{dw}{w^2+1}-\int\frac{dz}{1-z^2}\right)=-\frac{\sqrt{2}}{\sqrt{2^4}}\left(\arctan(w)-\text{arctanh}\right(z))+C$$
Note that $$w=\frac{u}{\sqrt{2}}=\frac{y-\frac{1}{y}}{\sqrt{2}}=\frac{\frac{1}{x}-x}{\sqrt{2}}$$ and $$z=\frac{t}{\sqrt{2}}=\frac{y+\frac{1}{y}}{\sqrt{2}}=\frac{\frac{1}{x}+x}{\sqrt{2}}$$ 
Thus $$I=\frac{1}{2\sqrt{2}}\left(\text{arctanh}\left(\frac{\frac{1}{x}+x}{\sqrt{2}}\right)-\text{arctan}\left(\frac{\frac{1}{x}-x}{\sqrt{2}}\right)\right)+C$$
