If $a, b > 0$ and $ b \neq 1$ prove that $\displaystyle {\int_1^b a^{\log_b x} dx > \ln b}$ Now, $a^{\log_b x} = x^{\log_b a},$
therefore $$ {\int_1^b a^{\log_b x}dx} =  {\int_1^b x^{\log_b a}dx} =  {\frac {ab - 1}{\log_b ab}} =   {\frac {ab - 1}{\ln ab - \ln 1} \ln b} =  {\frac {\ln b}{c}}$$ where $c \in (1 , ab)$ when $ab > 1$ or $c \in (ab , 1)$.
Here, I have proved the inequality for $c \in (1 , ab)$ as tangent of $\log x $ at $x>1$ is less than tangent at $x=1$. But I can not prove it when $c \in (ab , 1)$.
Is there any other easy approach to this problem?
 A: Consider the inequality
$$\tag1
\frac{ab-1}{\ln ab}>1,
$$
for all $a,b>0$. 
This is equivalent to 
$$\tag2
ab>1+\ln ab.
$$
The function $g(t)=t-1-\ln t$ has derivative $g'(t)=1-\tfrac1t$, so $g$ is decreasing on $(0,1)$ and increasing on $(1,\infty)$, with the only critical point, a minimum, at $t=1$. Thus $g(t)\geq g(1)=0$, and the inequality is strict when $t\ne1$. In summary, $(1)$ holds as long as $ab\ne1$, in which case you have equality and not strict inequality. 
Now, when $b<1$, we have that $\ln b<0$, and so the inequality that holds is 
$$
\int_1^b a^{\log_b x} < \ln b.
$$
A: Your statement doesn't hold as generally as you say it does. Consider cases:
Case 1: $ab<1,b<1$. Then $\frac{ab-1}{\ln(ab)} \ln(b) > \ln(b) \Leftrightarrow \frac{ab-1}{\ln(ab)}<1 \Leftrightarrow ab-1>\ln(ab)$. This holds because $g(t)=t-1-\ln(t)$ has a minimum value of $0$ at $t=1$.
Case 2: $ab<1,b>1$. Then $\frac{ab-1}{\ln(ab)} \ln(b) > \ln(b) \Leftrightarrow \frac{ab-1}{\ln(ab)}>1 \Leftrightarrow ab-1<\ln(ab)$. This does not hold for the same reason mentioned in case 1.
Case 3: $ab>1,b>1$. Then $\frac{ab-1}{\ln(ab)} \ln(b) > \ln(b) \Leftrightarrow \frac{ab-1}{\ln(ab)}>1 \Leftrightarrow ab-1>\ln(ab)$. This does hold for the reason mentioned in case 1.
For an example of case 2, consider $b=e,a=e^{-2}$, then $\frac{ab-1}{\ln(ab)} \ln(b)=1-e^{-1}<1=\ln(b)$.
