Lösung 2.2:3c
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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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 | ||
+ | {{Displayed math||<math>\int \ln x\cdot\frac{1}{x}\,dx\,,</math>}} | ||
- | <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 |
- | + | {{Displayed math||<math>\int u\cdot u'\,dx\,\textrm{.}</math>}} | |
- | + | ||
- | + | ||
- | + | ||
- | <math>u | + | |
+ | Thus, it seems that <math>u=\ln x</math> is a useful substitution, | ||
- | + | {{Displayed math||<math>\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] | |
- | <math>\begin{align} | + | &= \frac{1}{2}(\ln x)^2 + C\,\textrm{.} |
- | + | \end{align}</math>}} | |
- | u=\ln x | + | |
- | du= | + | |
- | \end{ | + | |
- | & =\int | + | |
- | & =\frac{1}{2} | + | |
- | \end{align}</math> | + |
Version vom 14:23, 28. Okt. 2008
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} |