Answer
\begin{aligned}
\frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} \ dt =\log _{a} x, \ \ \ x>0
\end{aligned}
Work Step by Step
Given $$\frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} \ dt $$
So, we have
\begin{aligned}
I&=\frac{1}{\ln a} \int_{1}^{x} \frac{1}{t} \ dt\\
&=\left[\frac{1}{\ln a} \ln |t|\right]_{1}^{x}\\
&=\frac{\ln x}{\ln a}-\frac{\ln 1}{\ln a}\\
\end{aligned}
Since $\frac{\ln x}{\ln a}=\log _{a}x$
We know that: $\ln 1 =0$
Thus:
$I=\log _{a} x, \ \ \ x>0$
