From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	It is simpler to investigate the integral if we write it as
-	<center> [[Image:2_2_3c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\int \ln x\cdot\frac{1}{x}\,dx\,,</math>}}
		+
		+	The derivative of <math>\ln x</math> is <math>1/x</math>, so if we choose <math>u = \ln x</math>, the integral can be expressed as
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		+	{{Displayed math||<math>\int u\cdot u'\,dx\,\textrm{.}</math>}}
		+
		+	Thus, it seems that <math>u=\ln x</math> is a useful substitution,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int \ln x\cdot\frac{1}{x}\,dx
		+	&= \left\{\begin{align}
		+	u &= \ln x\\[5pt]
		+	du &= (\ln x)'\,dx = (1/x)\,dx
		+	\end{align}\right\}\\[5pt]
		+	&= \int u\,du\\[5pt]
		+	&= \frac{1}{2}u^{2} + C\\[5pt]
		+	&= \frac{1}{2}(\ln x)^2 + C\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

It is simpler to investigate the integral if we write it as

\displaystyle \int \ln x\cdot\frac{1}{x}\,dx\,,

The derivative of \displaystyle \ln x is \displaystyle 1/x, so if we choose \displaystyle u = \ln x, the integral can be expressed as

\displaystyle \int u\cdot u'\,dx\,\textrm{.}

Thus, it seems that \displaystyle u=\ln x is a useful substitution,

\displaystyle \begin{align} \int \ln x\cdot\frac{1}{x}\,dx &= \left\{\begin{align} u &= \ln x\\[5pt] du &= (\ln x)'\,dx = (1/x)\,dx \end{align}\right\}\\[5pt] &= \int u\,du\\[5pt] &= \frac{1}{2}u^{2} + C\\[5pt] &= \frac{1}{2}(\ln x)^2 + C\,\textrm{.} \end{align}