Lösung 2.2:3c

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

Version vom 11:02, 21. Okt. 2008

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


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

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


\displaystyle \int{u\centerdot {u}'\,dx}


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


\displaystyle \begin{align} & \int{\ln x\centerdot \frac{1}{x}\,dx}=\left\{ \begin{matrix} u=\ln x \\ du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx \\ \end{matrix} \right\} \\ & =\int{u\,du=\frac{1}{2}u^{2}+C} \\ & =\frac{1}{2}\left( \ln x \right)^{2}+C \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3c“