Can you tell me whether my approach is correct.
We first switch to polar coordinates and we get the following integral:
$\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\infty}\frac{r}{r^{10}+5}d\phi dr=\int\limits _{0}^{\frac{\pi}{2}}d\phi\int\limits _{0}^{\infty}\frac{r}{r^{10}+5}dr=\frac{\pi}{2}\int\limits _{0}^{\infty}\frac{r}{r^{10}+5}dr$
From now on we solve it as it were an improper integral of single variable. We know that $\int\limits _{1}^{+\infty}\frac{1}{r^{\alpha}}$ is convergent $\iff\alpha\ge2$.
We use the following criteria. Let $F(x)\ge 0,\,G(x)\ge 0$ be such that $\lim\limits_{x\to\infty}\frac{F(x)}{G(x)}=\lambda,\lambda\in\mathbb{{R}}\cup\{+\infty\}$. If $\lambda$ is real and if $\int\limits_{a}^{+\infty}G(x)dx$ is convergent $\implies \int\limits _{a}^{+\infty}F(x)dx$ is also convergent.
We take $G(r)=\frac{1}{r^{9}}$ and compute $\lim\limits_{r\to\infty}\frac{F(r)}{G(r)}=\frac{r^{10}}{r^{10}+5}=1$, hence the integral is convergent.
It's simpler to say that $0 < \frac{r}{r^{10}+5} < \frac1{r^9}$ in $[1,\infty)$, and $\int\limits_1^{\infty}\frac1{r^9}dr$ converges, so $\int\limits_1^{\infty}\frac{r}{r^{10}+5}dr$ converges.
(Also, $0 \le \frac{r}{r^{10}+5} < \frac15$ in $[0,1]$, so $\int\limits_0^1\frac{r}{r^{10}+5}dr$ converges.)