The shortest way to prove that $\int_1^\infty \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx $ converges. I'm trying to show that the integral $$\int_1^\infty  \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx \quad \text{is convergent}.$$ 
We know that $$\frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }} < \frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}{\text{  for}}\,\,{\text{all}}\,\,\,x \in \left\langle {1, + \infty } \right\rangle $$
Now this is reduced to prove that $$\int_1^\infty  {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx $$ is convergent,
but the latter integral is difficult to calculate or prove to be convergent.
I know that this integral is convergent because when I calculate with Maple is
$$\int_1^\infty  {\frac{{\sqrt x }}{{\sqrt {{x^4} - 1} }}}dx  = \frac{{{\pi ^{3/2}}\csc \left( {\frac{\pi }{8}} \right)}}{{4\Gamma \left( {\frac{7}{8}} \right)\Gamma \left( {\frac{5}{8}} \right)}} \approx {\text{2}}{\text{.327185143}}$$
So my question is:
Is there a simpler way of proving that $\int_1^\infty  \frac{{\arctan \left( x \right)}}{{\sqrt {{x^4} - 1} }}dx$ converges?
 A: On $(1,+\infty)$ the function $\arctan x $ is positive and bounded by $\frac{\pi}{2}$, while:
$$ \int_{1}^{+\infty}\frac{dx}{\sqrt{x^4-1}}=\int_{0}^{1}\frac{dy}{\sqrt{1-y^4}}\leq\int_{0}^{1}\frac{dy}{\sqrt{1-y^2}}=\frac{\pi}{2}.$$
A: Note that your integral has infinities in two points: in $1$ because of the denominator, and in $\infty$. Split your integral in two pieces $\int \limits _1 ^2 + \int \limits _2 ^\infty$ and let us study each of them individually.
For the second one, let us apply the ratio test: call your function to integrate $f$, and let $g = \frac 1 {x^2}$. Note that $\lim \limits _{x \to \infty} \frac {f(x)} {g(x)} = \frac \pi 2 \in (0, \infty)$ which shows that $\int _2 ^\infty f \mathbb d x$ and $\int _2 ^\infty g \mathbb d x$ will behave similarly. But the second one is clearly convergent.
For the first one we shall use the ratio test again: consider the function $g(x)=\frac 1 {\sqrt {x-1}}$. Then $\lim \limits _{x \to 1^+} \frac {f(x)} {g(x)}= {\frac \pi 8} \in (0,\infty)$, which again shows that $\int _1 ^2 f \mathbb d x$ and $\int _1 ^2 g \mathbb d x$ will behave similarly. Again, the second integral is convergent.
So, the originl integral is convergent. The proof is quite short, it seems long because I have written down plenty of words and details.
A: The integral is improper at $x=1$ and $x=\infty$. We first split the integral at a point, say, $x=2$. 
For $\displaystyle\int_2^\infty\frac{dx}{\sqrt{x^4-1}}$, apply the limit comparison test to the integrand $\frac{1}{x^2}$, whose integral over $[2, \infty)$ is convergent by p-test.
For $\displaystyle\int_1^2\frac{dx}{\sqrt{x^4-1}}$, apply the substitution $x=1+y$, and we have the integral $\displaystyle\int_0^1\frac{dy}{\sqrt{4y+6y^2+4y^3+y^4}}$. Apply limit comparison test to the integrand $\frac{1}{\sqrt{y}}$, whose integral over $(1, 2)$ is convergent by p-test.
