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. 
 A: Your inequality isn't true for all $a,b>0$ with $b\neq 1$. Here's a proof for the domain where it is true:
$$\{(a,b)\in\mathbb{R}^+:ab>1\land b>1\}\cup\{(a,b)\in\mathbb{R}^+:ab<1\land b<1\}$$
The key is that this is the region in  $\frac{\ln(b)}{\ln(ab)}>0$ So we can multiply and divide inequalities by that without changing the sign. You have that you want to prove that
$$\frac{ab-1}{\ln(ab)}\ln(b)>\ln(b)$$
Since $\frac{\ln(b)}{\ln(ab)}>0$ we can divide both sides of the I equality by that. This gives $ab>1+\ln(ab)$ or, making the substitution $x=ab$, $x>1+\ln(x)$ and then exponentiating gives $e^x>ex\Rightarrow e^{x-1}>x$. Take the power series for $e^z$ and plug in $x-1$, to get
$$1+(x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots>x$$
Simplifying gives $$\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots>0$$
Which holds for all $x$ because it's either a strictly positive series or an alternating series that doesn't converge to zero with positive first term. All of these implications are if and only if statements, so the original inequality holds.
Outside of the specified region, the ratio $\frac{\ln(b)}{\ln(ab)}$ is negative, and so following this exact same line of reasoning proves that the reverse inequality is true outside of the specified region. The only case not addressed here is when $b=1$, in which case the inequality is false and instead equality holds.
A: 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
$$
\ln b\frac{ab-1}{\ln(ab)}-\ln b\approx -0.05.
$$
A: 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.
