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Lösung 2.3:1d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page)
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We can discern two factors in the integrand,
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<center> [[Image:2_3_1d.gif]] </center>
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<math>x</math>
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and
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<math>\ln x</math>. If we are thinking about using partial integration, then one factor should be integrated and the other differentiated. It can seem attractive to choose to differentiate
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<math>x</math>
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because then it will become equal to 1, but then we have the problem of determining a primitive function for
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<math>\ln x</math>
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(how is that done?). Instead, a more successful way is to integrate
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<math>x</math>
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and to differentiate
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<math>\ln x</math>,
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<math>\begin{align}
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& \int{x\ln x\,dx=\frac{x^{2}}{2}\ln x}-\int{\frac{x^{2}}{2}}\centerdot \frac{1}{x}\,dx \\
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& =\frac{x^{2}}{2}\ln x-\frac{1}{2}\int{x\,dx} \\
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& =\frac{x^{2}}{2}\ln x-\frac{1}{2}\centerdot \frac{x^{2}}{2}+C \\
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& =\frac{x^{2}}{2}\left( \ln x-\frac{1}{2} \right)+C \\
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\end{align}</math>
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Thus, how one should the factors in a partial integration is very dependent on the situation and there are no simple rules.

Version vom 13:49, 21. Okt. 2008

We can discern two factors in the integrand, x and lnx. If we are thinking about using partial integration, then one factor should be integrated and the other differentiated. It can seem attractive to choose to differentiate x because then it will become equal to 1, but then we have the problem of determining a primitive function for lnx (how is that done?). Instead, a more successful way is to integrate x and to differentiate lnx,


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


Thus, how one should the factors in a partial integration is very dependent on the situation and there are no simple rules.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:1d“