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Lösung 2.2:3c

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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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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{{Abgesetzte Formel||<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
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>}}
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{{Abgesetzte Formel||<math>\int u\cdot u'\,dx\,\textrm{.}</math>}}
Thus, it seems that <math>u=\ln x</math> is a useful substitution,
Thus, it seems that <math>u=\ln x</math> is a useful substitution,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int \ln x\cdot\frac{1}{x}\,dx
\int \ln x\cdot\frac{1}{x}\,dx
&= \left\{\begin{align}
&= \left\{\begin{align}

Version vom 13:02, 10. Mär. 2009

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“