<s>[INST] Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (something trival):
$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$
$$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
so it remains to prove that 
$$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
Thanks in advance!
 [/INST] We could use Carleman's Inequality which states that,
$$
\int_0^\infty f(x)dx \ge 1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx
$$
Then, by substituting our function as f(x), we can prove the statement by proving that 
$$
1/e\int_0^\infty exp(1/x*\int_0^xln(f(x))dx)dx \gt 1
$$
This is done by noting that for all x >0
$$
9t^3 + 9 \gt 8t^3 + t + 7
$$
Simplifying the right hand integral might prove the case.
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