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.)
On
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.
On
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.
On
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.
On
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.
On
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.
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*}