Definite Integral Using Identity (Boas 14.7.32) I need a little help in solving the following integral using a particular method: 

$$\int_0^\infty \frac {dx}{(1+x^4)^2}$$

The problem asks for a substitution of $x^4$ for $u$ and to use the following integral, 
$$\int_0^\infty \frac {r^{p-1}}{1+r}dr= \frac \pi{\sin{\pi p}}$$
which is true for $0<p<1$ .  A proof of this identity can be found here.  
It's pretty clear that my issue with this problem is the extra power of two.  I believe that the solution should be as simple as changing the simple pole to a second order one and using the same contour, but I could be wrong.  If that the case, the residue at $z=-1=e^{i\pi}$ would be
$$R(-1)=\frac 1 {1!} \frac d {dz} z^{p-1} \bigg|_{z=e^{i\pi}}=\frac{z^{p-2}}{p-1}\bigg|_{z=e^{i\pi}}= \frac {e^{i\pi}}{p-1}$$
and following from the Cauchy Integral formula we would have
$$\int_0^\infty \frac {r^{p-1}}{(1+r)^2}dr= -\frac {\pi}{(p-1)\sin{\pi p}}$$
Now, going back to the original problem and making the appropriate substitutions we find, 
$$\int_0^\infty \frac {dx}{(1+x^4)^2}=\frac 14\int_0^\infty \frac {u^{-3/4}}{(1+u)^2}du =-\frac 14 \frac {\pi}{(-3/4)\sin{(\pi /4)}}=\frac{\sqrt{2} \pi}3$$
Using Wolfram I get $\frac {3\pi}{8\sqrt2}$ for the actual answer. 
Since I'm stuck, any help as to what I did wrong or how to actually get the right answer would be greatly appreciated. 
*Found the obvious mistake in that I apparently forgot how to take a derivative.
 A: $\textbf{an idea}$
$$
\int \frac{1}{\left(1+x^4\right)^2}dx
$$
let $r = x^4$
thus we find
$$
\frac{1}{4}\int \frac{r^{-3/4}}{\left(1+r\right)^2}dr
$$
we can write
$$
\dfrac{d}{dr}\frac{r^{-3/4}}{\left(1+r\right)} = -\frac{r^{-3/4}}{\left(1+r\right)^2} -\frac{3}{4}\frac{r^{-7/4}}{\left(1+r\right)}
$$
or
$$
\frac{r^{-3/4}}{\left(1+r\right)^2} = -\frac{3}{4}\frac{r^{-7/4}}{\left(1+r\right)} - \dfrac{d}{dr}\frac{r^{-3/4}}{\left(1+r\right)} 
$$
thus
$$
\frac{1}{4}\int \frac{r^{-3/4}}{\left(1+r\right)^2}dr = \frac{1}{4}\int \left[-\frac{3}{4}\frac{r^{-7/4}}{\left(1+r\right)} - \dfrac{d}{dr}\frac{r^{-3/4}}{\left(1+r\right)} \right]dr
$$
or
$$
\int \frac{1}{\left(1+x^4\right)^2}dx =-\frac{1}{4}\int\dfrac{d}{dr}\frac{r^{-3/4}}{\left(1+r\right)}  dr  -\frac{3}{16}\int \frac{r^{-7/4}}{\left(1+r\right)}dr
$$
now you can see that the first integral on the RHS is straight forward and the second you have a formula for.
the result of 
$$
-\frac{3}{16}\int \frac{r^{-7/4}}{\left(1+r\right)}dr = \frac{3\pi}{8\sqrt{2}}
$$
$\textbf{There is an issue with integrating the first integral with the limits so it suggests that there is something missing}$
A: I found the mistake, but I'm going to generalize this problem for the sake of completeness. 
For any integral in the form: 
$$\int_0^\infty \frac{r^{p-1}}{(1+r)^n}$$
The residue at $-1=e^{i\pi}$ is
$$R(-1)=\frac 1 {(n-1)!} \frac {d^{n-1}} {dz^{n-1}} z^{p-1} \bigg|_{z=e^{i\pi}}=\frac{(p-1)(p-2)...(p-(n-1))}{(n-1)!}z^{p-n}\bigg|_{z=e^{i\pi}}$$
$$R(-1)=\frac{(p-1)(p-2)...(p-(n-1))}{(n-1)!}(-1)^{n-1} e^{i\pi}$$
Which means that, following the same logic as before, 
$$\int_0^\infty \frac{r^{p-1}}{(1+r)^n}=(-1)^{n-1}\frac {\pi}{\sin{\pi p}} \frac{(p-1)(p-2)...(p-(n-1))}{(n-1)!}$$
So for the $n=2$ case we would have
$$\int_0^\infty \frac{r^{p-1}}{(1+r)^n}=-\frac {\pi}{\sin{\pi p}}(p-1)$$
Now, solving for the particular problem
$$\int_0^\infty \frac {dx}{(1+x^4)^2}=\frac 14\int_0^\infty \frac {u^{-3/4}}{(1+u)^2}du =-\frac 14 \frac {\pi}{\sin{(\pi /4)}}(-3/4)=\frac{3\pi}{8\sqrt2}$$
