<s>[INST] 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...
 [/INST] Since$$16-x^4=(2-x)(8+4x+2x^2+x^3),$$you have$$\frac x{(16-x^4)^{1/3}}=\frac x{\sqrt[3]{2-x}\sqrt[3]{8+4x+2x^2+x^3}}$$and therefore$$\lim_{x\to2^-}\frac{\frac x{(16-x^4)^{1/3}}}{\frac1{\sqrt[3]{2-x}}}=\frac1{\sqrt[3]4}\ne0.$$So, since the integral$$\int_0^2\frac1{\sqrt[3]{2-x}}\,\mathrm dx$$converges, then so does your integral.
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