Solution 2.2:3c
From Förberedande kurs i matematik 2
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| - | {{ | + | It is simpler to investigate the integral if we write it as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<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 | ||
| + | |||
| + | {{Displayed math||<math>\int u\cdot u'\,dx\,\textrm{.}</math>}} | ||
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| + | Thus, it seems that <math>u=\ln x</math> is a useful substitution, | ||
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| + | {{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] | ||
| + | &= \frac{1}{2}(\ln x)^2 + C\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
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} |
