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.

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:
$$
x^4+1 = (x^2-x\sqrt2+1)(x^2+x\sqrt2 + 1).
$$
