Convergence of the improper integral $\int_0^2\frac x{(16-x^4)^{1/3}}\,\mathrm dx$ Need your help in the checking convergence of this improper integral :$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}$$I tried my luck and somehow was able to establish a relationship between  $(1-t^4)$  and $t^4$ i.e $$(1-t^4)\geq t^4$$
$$\frac{1}{1-t^4} \leq \frac{1}{t^4}$$
$$\Rightarrow \frac{1}{(1-t^4)\frac{1}{3}} \leq \frac{1}{t^\frac{4}{3}}$$ $$\Rightarrow \frac{t}{(1-t^4)\frac{1}{3}} \leq \frac{t}{t^\frac{4}{3}}$$
for some $$t\in\left[0,\frac{1}{2^\frac{1}{4}}\right]$$ It can be shown that$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}=2^\frac{2}{3}\int_0^1\frac{tdt}{(1-t^4)^\frac{1}3} $$
Therefore ,
$$F(x) =\int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{(1-t^4)^\frac{1}3} \leq \int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{t^\frac{4}{3}} = g(x)$$ For this particular interval I was able to prove its convergence as $g(x)$ converges but I am unable to comment on the remaining interval. Please help and also provide some insight which might help me in future. Thanks in advance...

The problem with your approach, as you observed, is that the inequality $1-t^4\ge t^4$ doesn't hold near $t=1$, which is where the impropriety in the integral occurs. Instead it's better to argue along the following lines:
For $0\le x\le2$, we have
$$16-x^4=(2-x)(8+4x+2x^2+x^3)\ge8(2-x)$$
and thus
$${x\over(16-x^4)^{1/3}}\le{2\over2(2-x)^{1/3}}=(2-x)^{-1/3}$$
so that
$$\int_0^2{x\over(16-x^4)^{1/3}}\,dx\le\int_0^2(2-x)^{-1/3}\,dx=-{3\over2}(2-x)^{2/3}\big|_0^2={3\over2^{1/3}}$$
