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| author | Andreas Müller <andreas.mueller@ost.ch> | 2022-05-25 12:08:44 +0200 |
|---|---|---|
| committer | Andreas Müller <andreas.mueller@ost.ch> | 2022-05-25 12:08:44 +0200 |
| commit | 4197abc20216c15f11660d63549eb8b765f1c892 (patch) | |
| tree | 6ae4b24ad67943c51a20015ae24ddfb285e0d9da /buch/chapters/060-integral | |
| parent | typo (diff) | |
| download | SeminarSpezielleFunktionen-4197abc20216c15f11660d63549eb8b765f1c892.tar.gz SeminarSpezielleFunktionen-4197abc20216c15f11660d63549eb8b765f1c892.zip | |
typos
Diffstat (limited to '')
| -rw-r--r-- | buch/chapters/060-integral/sqrat.tex | 6 |
1 files changed, 3 insertions, 3 deletions
diff --git a/buch/chapters/060-integral/sqrat.tex b/buch/chapters/060-integral/sqrat.tex index f6838e5..ceb8650 100644 --- a/buch/chapters/060-integral/sqrat.tex +++ b/buch/chapters/060-integral/sqrat.tex @@ -337,7 +337,7 @@ Durch Ableitung der Funktion \[ F = -\frac{1}{\sqrt{a}}\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{y}}\biggr) +\frac{1}{\sqrt{a}}\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{a}}\biggr) \] kann man nachprüfen, dass $F$ eine Stammfunktion von $1/y$ ist, also @@ -345,7 +345,7 @@ also \int \frac{1}{y} = -\frac{1}{\sqrt{a}}\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{y}}\biggr). +\frac{1}{\sqrt{a}}\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{a}}\biggr). \end{equation} % @@ -458,7 +458,7 @@ Form = v_0 + C -\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{y}}\biggr) +\log\biggl(x+\frac{b}{2a}+\frac{y}{\sqrt{a}}\biggr) + \sum_{i=1}^n c_i \log v_i, |