Lösung 2.3:1d

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We can discern two factors in the integrand,
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We can discern two factors in the integrand, <math>x</math> and <math>\ln x</math>. If we are thinking about using integration by parts, then one factor should be integrated and the other differentiated. It can seem attractive to choose to differentiate <math>x</math> because then it will become equal to 1, but then we have the problem of determining a primitive function for <math>\ln x</math> (how is that done?). Instead, a more successful way is to integrate <math>x</math> and to differentiate <math>\ln x</math>,
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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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{{Displayed math||<math>\begin{align}
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\int x\cdot\ln x\,dx
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&= \frac{x^2}{2}\cdot\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx\\[5pt]
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&= \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx\\[5pt]
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&= \frac{x^2}{2}\ln x - \frac{1}{2}\cdot\frac{x^2}{2} + C\\[5pt]
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&= \frac{x^2}{2}\bigl(\ln x-\tfrac{1}{2}\bigr) + C\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Thus, how one should choose the factors in an integration by parts is very dependent on the situation and there are no simple rules.
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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.
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Version vom 08:27, 29. Okt. 2008

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

\displaystyle \begin{align}

\int x\cdot\ln x\,dx &= \frac{x^2}{2}\cdot\ln x - \int \frac{x^2}{2}\cdot\frac{1}{x}\,dx\\[5pt] &= \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx\\[5pt] &= \frac{x^2}{2}\ln x - \frac{1}{2}\cdot\frac{x^2}{2} + C\\[5pt] &= \frac{x^2}{2}\bigl(\ln x-\tfrac{1}{2}\bigr) + C\,\textrm{.} \end{align}

Thus, how one should choose the factors in an integration by parts is very dependent on the situation and there are no simple rules.

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