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. 

As every other part of the answer has been cross-examined, I'm going to show the integration and hopefully you guys can spot what's gone wrong.
$$\mathrm I = \int_{1}^{b}a^{\log _bx}dx $$
After the substitution $x=b^u \implies dx=b^u \mathrm{ln}(b)du$ which is a nice continuous and differentiable function, we reduce the integral to $$ \int_0^1 a^u \times b^u \mathrm{ln}(b)du $$ as $x=1 \implies u=0 $ and $ x=b \implies u=1 $
Simplifying to $$ \mathrm I = \mathrm{ln}(b) \int_{0}^{1} (ab)^udu $$
As $\frac{(ab^u)}{\mathrm{ln}(ab)}$ is a primitive of $(ab)^u$ (Assuming $ab 
\ne 1$)and is continuous and differentiable we can apply the Newton-Leibnitz formula to give us $$ \mathrm I = \frac{\ ab-1}{{\ln ab}}{\ln b} $$
So Its understandable why the case $ab=1$ is problematic. Note that evaluating the integral for the case $ab=1$ returns an equality, contradicting the conclusion of the problem
The other possibilities may stem from the fact that while $ ab-1 > \mathrm{ln}(ab) $ The statement $ \frac{\ ab-1}{{\ln ab}} >1 $ also requires $ab>1$
Similarly the statement $ \frac{\ ab-1}{{\ln ab}} >1 \implies \ln(b) \frac{\ ab-1}{{\ln ab}} >ln(b)$ iff $b>1$ 
I see no reason to think that any assumptions have been made during the integration to exclude the previous two cases, If you do please comment.
