Integral comparison test $$ \int_a^b \frac{x+1}{\sqrt[4]{x^5+x^2}} \,dx $$
I want to know this integral converges or not when $(a, b) = (0, 1), (1, \infty)$. I was thinking of using the comparison test, but I can't think of anything to compare it to.
Thanks in advance.

For $(a,b)=(0,1)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\le\frac{x+1}{\sqrt[4]{x^{2}}}=\frac{x+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}$$
and for $(a,b)=(1,\infty)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{2x^{5}}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}\frac{x}{x^{\frac{5}{4}}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}x^{1-\frac{5}{4}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}x^{-\frac{1}{4}}.$$
Notice that for $x$ near $0$ we have $$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\approx\frac{1}{\sqrt[4]{x^{2}}}=x^{-\frac{2}{4}}=x^{-\frac{1}{2}}$$
and that for large $x$ we have $$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\approx\frac{x}{x^{\frac{5}{4}}}=x^{-\frac{1}{4}}.$$ So these estimates are not unexpected.
