Antwort 2.2:3
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
(Ny sida: {| width="100%" cellspacing="10px" |a) |width="50%"|<math>-\cos x^2+C</math> |b) |width="50%"| <math>\displaystyle\frac{\sin^2x}{2}+C</math> |- |c) |width="50%"| <math>\frac{1}{2}(\ln x)^2+...) |
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| (Der Versionsvergleich bezieht 2 dazwischen liegende Versionen mit ein.) | |||
| Zeile 6: | Zeile 6: | ||
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|c) | |c) | ||
| - | |width="50%"| <math>\frac{1}{2}(\ln x)^2+C</math> | + | |width="50%"| <math>\frac{1}{2}\,\,(\ln x)^2+C</math> |
|d) | |d) | ||
|width="50%"| <math>\displaystyle\frac{1}{2}\ln\left(x^2+2x+2\right)+C</math> | |width="50%"| <math>\displaystyle\frac{1}{2}\ln\left(x^2+2x+2\right)+C</math> | ||
Aktuelle Version
| a) | \displaystyle -\cos x^2+C | b) | \displaystyle \displaystyle\frac{\sin^2x}{2}+C |
| c) | \displaystyle \frac{1}{2}\,\,(\ln x)^2+C | d) | \displaystyle \displaystyle\frac{1}{2}\ln\left(x^2+2x+2\right)+C |
| e) | \displaystyle \displaystyle\frac{3}{2}\ln\left(x^2+1\right)+C | f) | \displaystyle -2\cos\sqrt{x}+C |
