Range of $\int_3^4 \frac{dx}{(\ln(x))^{1/3}}$ Make sure that the inequalities $\frac{x}{e}> \ln(x)>1$ hold true for $x>e$ and show that the integral $$\int_3^4 \frac{dx}{(\ln(x))^{1/3}}$$ is less than unity but more than $0.92$.
If I have made sure that $\frac{x}{e}> \ln(x)>1$ hold true for $x>e$ how could I use it to prove second part? Could someone help me with this?

If you know that $\frac xe > \ln x > 1$ for $x > e$ then in particular it is valid for $3 \le x \le 4$. The monotonicity of the integral in turn leads to $$ \int_3^4 \frac{e^{1/3}}{x^{1/3}} \, dx \le \int_3^4 \frac{1}{(\ln x)^{1/3}} \, dx \le \int_3^4 1 \, dx.$$ This gives you $$\frac 32 e^{1/3} (4^{2/3} - 3^{2/3}) \le  \int_3^4 \frac{1}{(\ln x)^{1/3}} \, dx \le 1.$$
The term on the right is a bit larger than $.9205$ per Wolfram Alpha.
