integrating the natural logarithm

One of the students asked me the other day how to integrate $f(x)=ln(x)$ – it’s not part of their course, but I thought I would do it here.

We use integration by parts to integrate $ln(x)$

$\begin{equation*}\int u dv=uv-\int v du\end{equation}$

$\begin{equation*}\int ln(x) dx\end{equation}$

Let $u=ln(x)$ and $dv=1$ , then $du=\frac{1}{x}$ and $v=x$

$\begin{equation*}\int ln(x) dx = xln(x)-\int x\times \frac{dx}{x}\end{equation}$

$\begin{equation*}\int ln(x) dx = xln(x)-\int 1 dx\end{equation}$

$\begin{equation*}\int ln(x) dx = xln(x)-x+c\end{equation}$

What about $\int 5ln(\sqrt{x}) dx$

We can take advantage of log laws and the properties of integration.

$\begin{equation*}\int 5ln(\sqrt{x}) dx=5\int lnx^{\frac{1}{2}} dx=5\int \frac{1}{2}ln(x) dx=\frac{5}{2} \int ln(x) dx\end{equation}$

$\begin{equation*}\frac{5}{2} \int ln(x) dx=\frac{5}{2}(xlnx-x)+c\end{equation}$

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