\section{Primitive of Logarithm of x over x}
Tags: Primitives involving Logarithm Function

\begin{theorem}
:$\ds \int \frac {\ln x} x \rd x = \frac {\ln^2 x} 2 + C$
\end{theorem}

\begin{proof}
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \ln x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac 1 x
      | c = Derivative of $\ln x$
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \ln x
      | c = Primitive of Reciprocal
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\ln x} x \rd x
      | r = \ln x \ln x - \int \ln x \paren {\frac 1 x} \rd x + C
      | c = Integration by Parts
}}
{{eqn | r = \ln^2 x - \int \frac {\ln x} x \rd x + C
      | c = tidying
}}
{{eqn | ll= \leadsto
      | l = 2 \int \frac {\ln x} x \rd x
      | r = \ln^2 x + C
      | c = adding $\ds\int \frac {\ln x} x \rd x$ to both sides
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\ln x} x \rd x
      | r = \frac {\ln^2 x} 2 + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}

