We are given the function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, $f(x)=\frac{|x|^2}{1+|x|^a}$ and want to find values of $a\in[0,\infty)$ such that the integral $\int_{\mathbb{R}^n} \frac{|x|^4}{(1+|x|^a)^2}\mathrm{d}\mathcal{L}^n(x)$ is finite.
My progress so far is that I have divided the integral into two parts $\int_{B_1^{(n)}(0)}\frac{|x|^4}{(1+|x|^a)^2}\mathrm{d}\mathcal{L}^n(x)$ and $\int_{\mathbb{R}^n\setminus B_1^{(n)}(0)}\frac{|x|^4}{(1+|x|^a)^2}\mathrm{d}\mathcal{L}^n(x)$.
Here the first integral is obviously finite for all values of $a$ so we can focus on the second integral.
If we then use polar coordinates we get a finite constant times the integral $\int_{[1,\infty)}\frac{r^4}{(1+r^a)^2}\cdot r^{n-1}\mathrm{d}\mathcal{L}^1(r)$. From this point I am not sure what is the best way to proceed.
I guess one could just try to use another substitution and calculate the integral directly but I feel like there is a different and perhaps shorter way to find a solution.
Another thing I have tried is to use the fact that the integral is monotone so I can use $1+r^a\geq r^a$ to get $\int_{[1,\infty)}\frac{r^{4+n-1}}{(1+r^a)^2}\mathrm{d}\mathcal{L}^1(r)\leq\int_{[1,\infty)}r^{4+n-1-2a}\mathrm{d}\mathcal{L}^1(r)$.
If I then integrate the right-hand side I get $\infty\overset{!}{>}\lim\limits_{r\rightarrow\infty}\frac{1}{4+n-2a}r^{4+n-2a}-\frac{1}{4+n-2a}$.
So we then get $a>\frac{4+n}{2}$ but I guess that is not the optimal solution.
$\int_{[1,\infty)}\frac{r^4}{(1+r^a)^2}\cdot r^{n-1}\mathrm{d}\mathcal{L}^1(r) <\infty$ if and only if $\int_{[1,\infty)}\frac{r^4}{(r^a)^2}\cdot r^{n-1}\mathrm{d}\mathcal{L}^1(r) <\infty$ if and only if $\int_{[1,\infty)} r^{4-2a+n-1} \mathrm{d}\mathcal{L}^1(r) <\infty$ if and only if $4-2a+n<0$ or $a >2+\frac n 2$ .
For the first equivalence you can use the fact that $r^{2a} \leq (1+r^{a})^{2} \leq 4r^{2a}$ for all $r >1$.