To find the integral of logarithms, we will use integration by parts:
\[ \int u \ dv = uv - \int v \ du \]
What we are trying to find:
\[\int \log_b x \ dx\]
Let \(u= \log_b x\) and \(v = x\):
\[\begin{align} \int u \ dv &= \int \log_b x \ dx \\ uv - \int v \ du &= (\log_b x)x - \int x \ du \end{align}\]
Now let's try to find \(du\):
\[\begin{gathered} \frac{d}{dx} u = (x \ln{b})^{-1} \\ du = (x \ln{b})^{-1} dx \end{gathered}\]
Now let's replace \(du\) in our integral:
\[\begin{align} uv - \int v \ du &= (\log_b x)x - \int x \ \left( \frac{1}{x \ln{b}} dx \right) \\ &= (\log_b x)x - \int (\ln{b})^{-1} \ dx \end{align}\]
And this leads to our answer:
\[ x(\log_b x) - \frac{x}{\ln{b}} + C\]
We can rewrite this:
\[ x\frac{\ln x}{\ln b} - \frac{x}{\ln{b}} + C = \frac{x}{\ln b} (\ln x -1) + C\]