$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$ 
*

*$x \geq 0$

*$n \geq 0$

*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$

*$I_n= \int_{0}^{ + \infty}f_n(x) $

*What is the limit of $I_n$ ?

*What is an equivalent of $I_n$ ?


$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&=  \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1}   }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n }  \right] +   \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1  \\
I_1&=  \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx 
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} =  \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim -  \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?

Complex analysis is one possibility. Note that $I_1 = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} \ dx = J$. Consider the semi-circle contour $C$ in the upper half of the complex plane and note that there are two poles (at $z_1 = e^{\pi i/4}$ and $z_2 = e^{3\pi i/4}$):
$$ 2\pi i \left( \frac{1}{4z_1} + \frac{1}{4z_2} \right) = \oint_C \frac{z^2}{1 + z^4} \ dz = J + \int_{C_R} \frac{z^2}{1+z^4} \ dz $$
where $C_R: z = R e^{i \theta}$ with $\theta \in [0,\pi]$. You can show that $|\int_{C_R}| \to 0$ as $R \to \infty$ and simplifying the above expression you obtain
$$ J = \frac{\pi}{\sqrt 2} \implies I_1 = \frac{\pi}{2\sqrt 2} $$
