Proving the inequality $\int_{1}^{b} a^{\log _bx}dx$ > ${\ln b}$ Proving the inequality
$$\int_{1}^{b}a^{\log _bx}dx > {\ln b}  : a,b>0 , b\neq1$$
Solving the integral, I found the result below,
$$\frac{\ ab-1}{{\ln ab}}{\ln b}$$
I know that I need only prove that, this part $$\frac{\ ab-1}{{\ln ab}}$$ must be greater than $1$, which I am unable to proceed further.
I ran to a difficulty when, $a= \frac{1}{b}$ or $b= \frac{1}{a}$ as $b$ ${\neq1}$
Please help.
Any alternative way to prove the inequality will be valued. 

I leave this as an answer, since it is the only way for me to include an image. I think the problem in the question is invalid, and should be updated to be correct (or removed if the true problem in itself is as stated).
I let Mathematica plot the region where
$$
\ln b\frac{ab-1}{\ln(ab)}<\ln b.
$$
It is the blue domain below.

For example, for $a=1/2$ and $b=4/3$ one gets
$$
\ln b\frac{ab-1}{\ln(ab)}-\ln b\approx -0.05.
$$
