Lösung 2.2:3c

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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

\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}

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