Integration of $\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$ with the help of signal properties. 
Find the integral
  $$\int_{0}^{\infty}\frac{dw}{1+\left ( \frac{w}{B} \right )^4}$$
  where $B$ is a constant.

This integration i tried by normal method that gives the result $\frac{\pi B}{2\sqrt2}$ but it goes so lengthy like this

is there any signal properties that i use to solve it in simple steps?

 A: With substitution $x^4=t$ and again with $\dfrac{1}{1+t}=u$ we have
\begin{align}
I 
&= B\int_0^\infty\dfrac{1}{1+x^4}dx \\
&= \dfrac{B}{4}\int_0^\infty\dfrac{t^{-\frac34}}{1+t}dt \\
&= \dfrac{B}{4}\int_0^1 u^{-\frac14}(1-u)^{-\frac34}du\\
&= \dfrac{B}{4}\beta(\frac14,\frac34) \\
&= \dfrac{B}{4}\dfrac{\Gamma(\dfrac14)\Gamma(\dfrac34)}{\Gamma(1)} \\
&= \dfrac{B}{4}\dfrac{\pi}{\sin\frac14\pi} \\
&= \color{blue}{\dfrac{\pi B}{2\sqrt{2}}}
\end{align}
where $\beta(x,y)$ is beta function.
A: Just want to share a "smart" method.
(Too long for comment)
Here is a fast way to evaluate the indefinite  integral.
Sub $ t=Bx$,
Then, 
\begin{align*}
\int \frac 1{1+x^4}\,dx&=\frac 12\left(\int \frac {1+x^2}{1+x^4}\,dx+\int \frac {1-x^2}{1+x^4}\,dx\right)\\
&=\frac 12\left(\int \frac {\color{red}{\left(\frac 1 {x^2}+1\right)dx}}{\frac 1{x^2}+x^2}+\int \frac {\color{blue}{\left(\frac 1 {x^2}-1\right)dx}}{\frac 1{x^2}+x^2}\right)\\
&=\frac 12\left(\int \frac {\color{red}{d\left(x-\frac 1 x\right)}}{\left(x-\frac 1 x\right)^2+2}+\int \frac {\color{blue}{-d\left(x+\frac 1 x\right)}}{\left(x+\frac 1 x\right)^2-2}\right)\qquad\quad\\
&= \frac 12\left(\frac 1{\sqrt2} \tan^{-1}\left(\frac {x-\frac 1 x} {\sqrt2}\right)+\frac 1{\sqrt2} \tanh^{-1}\left(\frac {x+\frac 1 x} {\sqrt2}\right)\right).
\end{align*}
A: A way to do the integral over the whole half-line with the residue theorem is the following. Consider the contour going straight from $0$ to $R$, following a circular arc from $R$ to $iR$, and then going straight from $iR$ to $0$. The desired integral $I$ is the integral on the first part of the contour. The last piece is
$$\int_R^0 \frac{1}{(ix)^4+1} d(ix)=i \int_R^0 \frac{1}{x^4+1} dx = -iI.$$
The middle piece can be shown using an "ML estimate" to go to zero as $R \to \infty$. Now the only pole of the integrand inside the contour is at $e^{i\pi/4}$, so the residue theorem tells you that the integral around the contour is
$$J=(1-i)I=2\pi i\operatorname{Res} \left ( \frac{1}{z^4+1},e^{i \pi/4} \right ).$$
This residue is $\lim_{z \to e^{i\pi/4}} \frac{z-e^{i\pi/4}}{z^4+1}=\frac{1}{4} e^{-3 i \pi/4}$, so $I=\frac{\pi i}{2(1-i)} e^{-3 i \pi/4}=\frac{\pi}{2\sqrt{2}}$.
