Evaluating $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ I was evaluating

$$\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$$

My work:
I see that in the denominator, there is the radical $\sqrt{x^2 + 3}$. This reminds me of the trigonometric substitution $\sqrt{u^2 + a^2}$ and letting 
$ u = a \tan \theta$. With this in mind: I let $x = \sqrt{3} \tan \theta$ .Moving on, $dx = \sqrt{3} (\sec \theta )^2 $. And.....if $\theta$ is an 
acute angle, we see in the illustration below that...

$\tan \theta = \frac{x}{\sqrt{3}}$ 
Then...getting the equivalent form of $\sqrt{x^2 + 3}$ in terms of $\theta$:
$$\sqrt{x^2 + 3}$$
$$ =\sqrt{(\sqrt{3} \tan \theta)^2 + 3}$$
$$= \sqrt{3(\tan \theta)^2 + 3}$$
$$= \sqrt{3((\tan \theta)^2 + 1)}$$
$$= \sqrt{3} \sqrt{(\sec \theta)^2} $$
$$\sqrt{x^2 + 3} = \sqrt{3} \sec \theta $$
Getting the equivalent form of $x^4$ in terms of $\theta$:
$$x^4 = (\sqrt{3} \tan \theta)^4$$
$$ = 9 (\tan \theta)^4 $$
Getting the equivalent form of $dx$ in terms of $\theta$:
$$dx = \sqrt{3} (\sec \theta )^2 d\theta$$
Substituting these equivalent expressions to the integral above, we get:
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{1}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta$$
$$ = \int \frac{\sqrt{3} (\sec \theta )^2}{9 \sqrt{3} (\tan \theta)^4 \sec \theta } d\theta$$
$$ = \int \frac{\sec \theta}{9  (\tan \theta)^4 } d\theta$$
$$ = \int \frac{\sec \theta}{9  ((\tan \theta)^2)^2   } d\theta$$
$$ = \int \frac{\sec \theta}{9  ((\sec \theta)^2 - 1 )^2   } d\theta$$
$$ = \int \frac{\sec \theta}{9  ((\sec \theta)^2 - 1 )^2   } d\theta$$
$$ = \frac{1}{9}\int \frac{\sec \theta}{((\sec \theta)^2 - 1 )^2   } d\theta$$
$$\int \frac{1}{x^4 \sqrt{x^2 + 3} } dx = \frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$$
My problem is....I don't know how to evaluate $\frac{1}{9}\int \frac{\sec \theta}{ (\sec \theta)^4 -2(\sec \theta)^2 +1 } d\theta$. 
I can't go forward. I'm stuck. How to evaluate $\int^\infty _1 \frac{1}{x^4 \sqrt{x^2 + 3} } dx$ properly?

Hint: 
You have done a good substitution so here you are
$$\int \frac{1}{9 (\tan \theta)^4 \sqrt{3} \sec \theta } \sqrt{3} (\sec \theta )^2 d\theta=\dfrac{1}{9}\int\dfrac{\cos^2\theta}{\sin^4\theta}\cos\theta\,d\theta=\dfrac{1}{9}\int\dfrac{1-\sin^2\theta}{\sin^4\theta}\cos\theta\,d\theta$$
now let $\sin\theta=u$ and continue!
