Is $\int_0^1 \frac{x^n}{\sqrt{1-x^4}}dx$ convergent?

Question

$$\int_0^1 \frac{x^n}{\sqrt{1-x^4}}dx$$

Near $0$ the expression inside is convergent, that is easy. Near $1$ looks like it approaches infinity when $n \ge 0$ But according to the book when $n \ge -1$ the integral is convergent. No proof is given. I am having difficulty to figure out why and how.

I would like an interface for checking integrals convergence. There is no need to find the number to which it converges. Just check.

Answer 1

Near $x=1$, the integrand blows up like $\dfrac1{\sqrt{1-x}}$, so the integral converges there.

So the only real problem is near $x=0$, for which you need $n>-1$.

Answer 2

Might be helpful for someone to know this integral has a precise value as Beta function. With substitution $x^4=t$ it is $$\int_0^1 \frac{x^n}{\sqrt{1-x^4}}dx=\dfrac14\int_0^1 t^\frac{n-3}{4}(1-t)^\frac{-1}{2} \ dt = \dfrac14\beta\left(\dfrac{n+1}{4},\dfrac12\right)$$ valid for $n>-1$.