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Lösung 2.2:3c

Aus Online Mathematik Brückenkurs 2

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It is simpler to investigate the integral if we write it as
It is simpler to investigate the integral if we write it as
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{{Displayed math||<math>\int \ln x\cdot\frac{1}{x}\,dx\,,</math>}}
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<math>\int{\ln x\centerdot \frac{1}{x}\,dx}</math>,
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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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The derivative of
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{{Displayed math||<math>\int u\cdot u'\,dx\,\textrm{.}</math>}}
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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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Thus, it seems that <math>u=\ln x</math> is a useful substitution,
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<math>\int{u\centerdot {u}'\,dx}</math>
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{{Displayed math||<math>\begin{align}
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\int \ln x\cdot\frac{1}{x}\,dx
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&= \left\{\begin{align}
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Thus, it seems that
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u &= \ln x\\[5pt]
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<math>u=\ln x</math>
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du &= (\ln x)'\,dx = (1/x)\,dx
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is a useful substitution,
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\end{align}\right\}\\[5pt]
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&= \int u\,du\\[5pt]
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&= \frac{1}{2}u^{2} + C\\[5pt]
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<math>\begin{align}
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&= \frac{1}{2}(\ln x)^2 + C\,\textrm{.}
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& \int{\ln x\centerdot \frac{1}{x}\,dx}=\left\{ \begin{matrix}
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\end{align}</math>}}
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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>
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Version vom 14:23, 28. Okt. 2008

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

lnxx1dx 

The derivative of lnx is 1x, so if we choose u=lnx, the integral can be expressed as

uudx. 

Thus, it seems that u=lnx is a useful substitution,

lnxx1dx=udu=lnx=(lnx)dx=(1x)dx=udu=21u2+C=21(lnx)2+C.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:3c“