I know that $\int_1^\infty \frac1xdx$ diverges. I can probe that $\int_1^\infty \frac1{\ln(x)}dx$ diverges, as $\forall x>1:\frac1x<\frac1{\ln(x)}$
I also know that $\int_1^\infty \frac1{x^2}dx$ converges $ \therefore \int_1^\infty \frac1{x^7}dx$ also converges.
I know that $\forall x>1:x>\ln(x) \rightarrow x^7>\ln^7(x)\rightarrow \frac1{x^7}<\frac1{\ln^7(x)}$ but this doesn't help me much.
Since $\ln x = 7 \ln x^{1/7}$ we have $\ln ^7 x = (7 \ln x^{1/7})^7 < 7^7 x$ and
$$\int _2^\infty \frac{dx}{\ln^7 x} > \int _2^\infty \frac{dx}{7^7x}= +\infty$$
The integral over $[1,2]$ also diverges, which can be shown by the limit comparison test with $(x-1)^{-7}$.
Take $x = e^u$. Your integral transforms to $\int_0^\infty \frac{e^u}{u^7}du$. You can now compare this with $\int_0^\infty \frac{1}{u^7}du$, since $1 < e^u$ for all $u > 0$. Since $\int_0^\infty \frac{1}{u^7}du$ diverges, it follows that $\int_0^\infty \frac{e^u}{u^7}du$ must diverge via Direct Comparison.
hint Near infinity $$\lim_{x\to+\infty}x.\frac{1}{(\ln(x))^7}=+\infty\implies$$ for $ x $ large enough
$$x.\frac{1}{(\ln(x))^7}>1\implies$$
$$\frac{1}{(\ln(x))^7}>\frac 1x\implies $$ $$\int_1^{+\infty}\frac{dx}{(\ln(x))^7}\text{ diverges}$$
OR
Near $ 1^+$
$$\ln(x)\sim (x-1)\implies $$ the two integrales $\int_1^2\frac{dx}{(\ln(x))^7}$ and $ \int_1^2\frac{dx}{(x-1)^7}$ have the same nature, divergent.
Nevermind, I made variable subtitution of $t=ln(x) \rightarrow e^tdt=dx$
So the integral ends up as $$\int_0^\infty \frac{e^t}{t^7}dt$$ But the limit of the integrand goes to infinity $$\lim_{t\to+\infty} \frac{e^t}{t^7}=\lim_{t\to+\infty} \frac{e^t}{7t^6}=...=\lim_{t\to+\infty} \frac{e^t}{7!}\to+\infty$$ $\therefore \int_0^\infty \frac{e^t}{t^7}dt$ diverges so $\int_1^\infty \frac1{ln^7(x)}dx$ diverges
