I want to investigate the convergence of integral
$$\int_0^1 \frac{1}{\sqrt[3]{- x^{4} + 1}}\, dx$$
How shall I deal with it? I thought about taking $\frac{1}{x^p}$ and finding $p$ by $\lim\limits_{x \to 1} \frac{f(x)}{g(x)}$. Now I decided to use $\frac{1}{(1-x)^p}$, but how shall I do with finding $p$?
And others: $\int\limits_1^2 \frac{1}{\ln x}\, dx$ $\int\limits_{n/2}^\infty \sin x\frac{1}{x^2}\, dx$
How should I deal with them, just an idea please
Let x=$\frac1y\implies dx=-\frac1{y^2}dy$ then
$$\int_{0}^{1} \frac{1}{\sqrt[3]{- x^{4} + 1}}\, dx=\int_{1}^{+\infty} \frac{\sqrt[3]y}{y\sqrt[3]{y^{4} - 1}}\, dy$$
and thus since
$$ \frac{\sqrt[3]y}{y\sqrt[3]{y^{4} - 1}}\sim \frac1{y^2}$$
the integral converges by limit comparison test with $\int \frac1{y^2}$.