Processing Math: Done
Lösung 2.3:1d
Aus Online Mathematik Brückenkurs 2
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- | We can discern two factors in the integrand, | + | 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>, |
- | <math>x</math> | + | |
- | and | + | |
- | <math>\ln x</math>. If we are thinking about using | + | |
- | <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>, | + | |
+ | {{Displayed math||<math>\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}</math>}} | ||
- | + | 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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- | Thus, how one should the factors in | + |
Version vom 08:27, 29. Okt. 2008
We can discern two factors in the integrand,
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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.