Is $\int_0^\infty\frac{\arctan x}{\sqrt{x^3+x}}dx$ converges/diverges

Question

I need to determine if the following integral converges/diverges.

$$
\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx
$$

What i tried:

We can write the integral as:

$$
\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx
$$

Because the $x^2+1$ i thought about defining $x = \tan u$, therefore i will be able to use the identity:
$$
\tan^2u + 1 = \frac{1}{\cos^2u}
$$

Define:
$$
x = \tan u \Rightarrow u = \arctan x
$$
$$
x = 0 \Rightarrow u = 0
$$
$$
x \to \infty \Rightarrow u = \pi/2
$$

Therefore we can write the integral as:

$$
\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx = \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u(\tan^2u+1)}}
$$

$$
= \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u\frac{1}{\cos^2u}}} = \int_{0}^{\pi/2}\frac{u\cos u}{\sqrt{\tan u}}du
$$

And here i am stuck.

Another way:

I thought maybe to devide the integral into two, not sure even how, maybe:

$$
\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\pi/2}\frac{\arctan x}{\sqrt{x^3+x}}dx + \int_{\pi/2}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx
$$

But i dont see how it gives me something.

Can i have a hint?

Thank you.

Answer 1

The long term behavior of the function is $\frac{\pi/2}{x^{3/2}}.$ The convergence of $\int_1^{\infty} x^{-3/2} \, dx$ should now tell you something.