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?


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*}
