Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$ finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$
My attempt is:
let $x=\sqrt{u}$
$dx=\frac{1}{2\sqrt{u}}$
$$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$
here I stopped because I don't know how to complete this solution. Any help please.
 A: One way to deal with this is to write
$$
1+x^4 = (1 + 2x^2+x^4) - 2x^2 = (1+x^2)^2 - (x\sqrt 2)^2 = (1+x^2+x\sqrt2)(1+x^2-x\sqrt 2)
$$
Then use partial fractions:
$$
\frac {Ax+B} {1+x^2+x\sqrt2} + \frac {Cx+D} {1+x^2-x\sqrt2}
$$
Notice that you can tell that these quadratic polynomials are irreducible because their discriminants $b^2-4ac = (\sqrt 2)^2 - 4\cdot1\cdot1$ are negative.  You have to do a bit of algebra to find $A$, $B$, $C$, and $D$.
If you let $u=x^2+x\sqrt2+1$ then $du=(2x+\sqrt2)\,dx$ and
$$
\frac{Ax+B}{x^2+x\sqrt2+1} \text{ becomes }\frac{\frac A 2 (2x+\sqrt 2)}{x^2+x\sqrt2+1} + \frac{\text{some consant}}{x^2+x\sqrt2+1}
$$
The substitution above handles the first fraction.  The second one is done by the substitution below:
\begin{align}
\int \frac {dx} {1+x^2+x\sqrt2} & = \int \frac{dx}{\left(x^2+x\sqrt2+\frac 1 2 \right) + \frac 1 2} \qquad(\text{completing the square}) \\[8pt]
& = \int \frac {dx}{\left(x + \frac 1 {\sqrt2}\right)^2 + \frac 1 2} = \int \frac{2\,dx}{\left(x\sqrt2 + 1\right)^2 + 1}.
\end{align}
Then use the substitution
\begin{align}
x\sqrt2+1 & = \tan\theta \\[6pt]
\text{so that }\left(x\sqrt2+1\right)^2 + 1 & = \sec^2\theta \\[6pt]
\text{and }\sqrt 2\,dx & = \sec^2\theta\,d\theta
\end{align}
and the integral above becomes
$$
\int \sqrt2\,d\theta = \sqrt 2\ \theta = \sqrt 2\ \arctan(x\sqrt2 + 1) +C
$$
etc.
Perhaps I should add that the first time I ever saw this integral I did it by a more straightforward method and consequently saw how to do what I did above, but some people don't like that more straightforward method because it involves complex numbers.  That method is this: to factor $x^4+1$ we solve $x^4+1=0$, getting
\begin{align}
x^4 & = -1 \\[6pt]
x^2 & = \pm i
\end{align}
To find square roots of $i$, notice that $i = \cos90^\circ+i\sin90^\circ$, so its square roots must be $\pm(\cos45^\circ+i\sin45^\circ)= \pm\left(\frac{1+i}{\sqrt2}\right)$ and similarly square roots of $-i$ are $\pm(\cos45^\circ-i\sin45^\circ)= \pm\left(\frac{1-i}{\sqrt2}\right)$  Now we have
$$
x^4+1=\underbrace{\left(x-\frac{1+i}{\sqrt2}\right)\left(x-\frac{1-i}{\sqrt2}\right)}_\text{conjugates}\  \underbrace{\left(x-\frac{-1+i}{\sqrt2}\right)\left(x-\frac{-1-i}{\sqrt2}\right)}_\text{conjugates}.
$$
When conjugates are multiplied, the imaginary parts cancel out:
$$
x^4+1 = (x^2-x\sqrt2+1)(x^2+x\sqrt2 + 1).
$$
A: $$\int_0^\infty\frac{dx}{1+x^4}=\Re\left({\int_0^\infty\frac1{1-ix^2}\,dx}\right).$$
Then by decomposition ,
$$\int_0^\infty\frac{dx}{1-ix^2}=\frac12\int_0^\infty\left(\frac1{1-\sqrt ix}+\frac1{1+\sqrt ix}\right)\,dx=\frac1{2\sqrt i}\ln\left(\frac{1+\sqrt ix}{1-\sqrt ix}\right)\Big|_0^\infty=\frac{i\pi-0}{2\sqrt i},$$
$$\color{green}{I=\Re\left(\frac{\sqrt i\pi}2\right)=\frac\pi{\sqrt8}}.$$
A: $$\int_{0}^{\infty }\frac{1}{1+x^4}dx=$$
$$\int_{0}^{\infty }\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right)dx$$
The antiderivative of $\left(\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)}\right) $ is:
$$\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))$$
So we got the limit of b goes to infinity:
$$\left[\frac{1}{4\sqrt{2}}(-\log(x^2-\sqrt{2}x+1)+\log(x^2+\sqrt{2}x+1)-2\tan^{-1}(1-\sqrt{2}x)+2\tan^{-1}(\sqrt{2}x+1))\right]^b_0$$
So we get:
$$\int_{0}^{\infty }\frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}$$
A: Method 1:
Because we have an even integrand, we can rewrite
$$
I=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{1+x^4}
$$
Now we are ready to use residue theorem. 
A close inspection of the denominator shows that there are 2 poles in the upper half plane :
$$
x_{\pm}=\frac{\pm1+i}{\sqrt{2}}
$$
Choosing a big semicircle in the uhp as an integration contour(the arc will vanish because the the integrand falls of as $1/R^3$ as $R\rightarrow \infty$ ) we get
$$
I= \frac{1}{2}\times2 \pi i \times[\text{Res}(x_+)+\text{Res}(x_-)]=\frac{\pi}{2\sqrt{2}}
$$
Method 2:
Another (maybe too) sophisticated approach would be as follows
Write $x^4=q, dx= \frac{1}{4}q^{-3/4}dq$ and we get 
$$
I=\frac{1}{4}\int_{0}^{\infty}\frac{1}{(1+x)x^{3/4}}
$$
which yields 
$$
I=\frac{1}{4}B(\frac{1}{4},\frac{3}{4})
$$
where we used one of the representations of Euler's Beta function. Using the fact the reflection formula for the Gamma function ($\Gamma(1-x)\Gamma(x)=\frac{\pi}{\sin(\pi x)}$) it follows that 
$$
I= \frac{\pi}{\sin(3\pi/ 4)}= \frac{\pi}{2 \sqrt{2}}
$$
A: Here's an approachable method without complex analysis.
Notice that you can write your integral as (which i leave to you why?)
$$\frac{1}{2}\int \frac{1+x^2}{1+x^4}\cdot dx+\frac{1}{2}\int \frac{1-x^2}{1+x^4}\cdot dx $$
I will only solve one of these and the other one for you.
Let me take 
$$\int \frac{1-x^2}{1+x^4}\cdot dx$$
Taking $x$ common,
$$\int \frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+x^2}\cdot dx $$
Now use the substitution $x-\frac{1}{x}=t$
Find $dt$ and also square the substitution to see what you can do for the denominator. 
$$\left(1-\frac{1}{x^2}\right))\,dx=dt$$
and $x^2+\frac{1}{x^2}=t^2+2$
Gives the integral 
$$\int \frac{-dt}{t^2+2}$$
This is an trivial integral of $\tan^{-1}$ 
And the other one i suspect is gonna be of the form $\frac{c}{x^2-a^2}$ probably a log but i leave that to you to try.
:) 
