I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
Many duplicate posts link to this one as the target. (Those posts were merged into this one, which is the source of the many answers.)
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}}{\left(x+\frac{1}{x}\right)^{2} - 2} \ dx + \text{same trick} \end{align*}
$$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$
$$\int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{\left(x-\frac1x\right)^2+2}dx$$
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
My hint:
\begin{align} \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{1}{2\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right]+\text{Constant} \end{align}
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 $1/(x^2+1)$.
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.
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.
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*}
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 $x$.
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:
$$ \int \frac{1}{x^4 +1} \ dx = \int \frac{\frac{x}{2\sqrt2}+\frac{1}{2}}{x^2+\sqrt2x+1} \ dx+\int \frac{\frac{-x}{2\sqrt2}+\frac{1}{2}}{x^2-\sqrt2x+1} \ dx=$$
$$ \frac{\sqrt2}{8} \int \frac{2x+2\sqrt2}{x^2+\sqrt2x+1} dx-\frac{\sqrt2}{8} \int \frac{2x-2\sqrt2}{x^2-\sqrt2x+1} dx=$$
$$\frac{\sqrt2}{8} \int \frac{2x+\sqrt2}{x^2+\sqrt2x+1} dx +\frac{1}{4} \int \frac{1}{x^2+\sqrt2x+1} dx $$ \qquad-\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.$$
Q.E.D.
It is posiible to use faktorization $1+x^4=(1-\sqrt{2}x+x^2)(1+\sqrt{2}x+x^2)$ and partial fractions.
Hint
Use $$1+t^4 = (1 + \sqrt{2} t + t^2 ) \times (1 - \sqrt{2} t + t^2 )$$ and decompose in partial fractions. You will arrive to much simpler antiderivatives.
I am sure that you can take from here.
What Chandrasekhar wrote is a very nice trick. I'll offer you here a more "standard" one: $$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} $$and now do partial fractions and find the coefficients $\,A,B,C,D$
Added...or wait until someone else do it for you, of course.
There are two (three) ways to go. One, assume
$$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}}}} $$
However, Chandrasekar's is the best way to go, if you can figure it out.
The steps are as follows:
1) Decompose $\frac{1}{1+x^{4}}$ using partial fractions (it can be factored using an identity of Sophie Germain)
2) You should have a linear function in each numerator and a quadratic in each denominator. Separate into the form $\frac{const}{quadratic}+\frac{const\cdot x}{quadratic}$
3) Complete the square on this quadratic.
4) To integrate the first form, make a simple substitution to transform the integrand into the form $\frac{1}{1+u^{2}}$, which is the derivative of $\tan^{-1}(x)$.
5) For the second, make another substitution to transform the integrand into the form $\frac{1}{1+v}$, which has antiderivative $\ln(1+v)$.
Be very careful with tiny algebraic slips, and keep track of your constants.
Hints:
$$x^4+1=(x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)$$
Partial Fractions, and yes: you'll need arctangents.
HINT (for partial fractions) : $$ x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1). $$
You can use $$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\}.$$
