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authorAndreas Müller <andreas.mueller@ost.ch>2022-05-25 12:08:44 +0200
committerAndreas Müller <andreas.mueller@ost.ch>2022-05-25 12:08:44 +0200
commit4197abc20216c15f11660d63549eb8b765f1c892 (patch)
tree6ae4b24ad67943c51a20015ae24ddfb285e0d9da /buch/chapters
parenttypo (diff)
downloadSeminarSpezielleFunktionen-4197abc20216c15f11660d63549eb8b765f1c892.tar.gz
SeminarSpezielleFunktionen-4197abc20216c15f11660d63549eb8b765f1c892.zip
typos
Diffstat (limited to '')
-rw-r--r--buch/chapters/060-integral/sqrat.tex6
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,