+For LyX 2.2.1:
+
+- sec. 23.1: new note
+
+
+For LyX 2.2.0:
+
In the first step:
Modified:
-
- sec. 15.1: Spanish only: new note behind the table
- sec. 16.1: Spanish only: correct note in the table
- sec. 18.1: Japanese only: updated paragraph
New:
-
- sec. 13.3: new note behind the table
-Modified:
+For LyX 2.2.1:
+
+- sec. 6.11: new sentence
+
+
+For LyX 2.2.0:
in the first step:
- sec. 3.3.10.1 (Japanese only) changed sentence
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
-\author 5863208 "ab"
\end_header
\begin_body
\end_layout
\begin_layout Itemize
-
-\change_inserted 5863208 1465782906
\begin_inset Formula $\int\left(\frac{1}{1+x^{3}}\right)dx=-\frac{\log\left(x^{2}-x+1\right)}{6}+\frac{\arctan\left(\frac{2\,x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log\left(x+1\right)}{3}$
\end_inset
-\begin_inset Foot
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Note Greyedout
status open
\begin_layout Plain Layout
-\change_inserted 5863208 1465782906
-Note that one needs to use proper delimiter insets
+\series bold
+Note:
+\series default
+ One needs to use proper delimiter insets
\begin_inset Formula $\left(\right)$
\end_inset
instead of simple '(' ')' characters.
-
\end_layout
\end_inset
-
-\change_unchanged
-
+
\end_layout
\begin_layout Itemize
\end_layout
-\begin_layout Standard
-
-\change_inserted 5863208 1465782942
-One can also use standard commands known to CAS:
-\end_layout
-
\begin_layout Itemize
-
-\change_inserted 5863208 1465782942
-\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit i_{2}}=0}^{\infty}{\frac{4^{-{\mathit i_{2}}-1}\,\left(x-1\right)^{{\mathit i_{2}}+1}}{{\mathit i_{2}}+1}}-\log4$
+\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit{i}_{2}}=0}^{\infty}{\frac{4^{-{\mathit{i}_{2}}-1}\,\left(x-1\right)^{{\mathit{i}_{2}}+1}}{{\mathit{i}_{2}}+1}}-\log4$
\end_inset
\end_layout
\begin_layout Itemize
-
-\change_inserted 5863208 1465782942
\begin_inset Formula $solve\left(x_{1}+y_{1}^{3}=y_{1}+x_{1}^{2},x_{1}\right)=\left[x_{1}=-\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}-1}{2},x_{1}=\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}+1}{2}\right]$
\end_inset
-\change_unchanged
-
\end_layout
\begin_layout Subsection
fraction bar is round about three times the bar thickness.
\end_layout
+\begin_layout Standard
+\begin_inset Newpage newpage
+\end_inset
+
+
+\end_layout
+
\begin_layout Subsection
Canceled Formulas
\begin_inset Index idx
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\left(\frac{1}{1+x^{3}}\right)dx=-\frac{\log\left(x^{2}-x+1\right)}{6}+\frac{\arctan\left(\frac{2\,x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log\left(x+1\right)}{3}$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+Achtung:
+\series default
+ Es muss die Einfügung für automatische Klammern
+\begin_inset Formula $\left(\right)$
+\end_inset
+
+ verwendet werden statt einfacher zeichen für Klammern '(' ')'.
+\end_layout
+
+\end_inset
+
+
\end_layout
\begin_layout Itemize
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit{i}_{2}}=0}^{\infty}{\frac{4^{-{\mathit{i}_{2}}-1}\,\left(x-1\right)^{{\mathit{i}_{2}}+1}}{{\mathit{i}_{2}}+1}}-\log4$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $solve\left(x_{1}+y_{1}^{3}=y_{1}+x_{1}^{2},x_{1}\right)=\left[x_{1}=-\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}-1}{2},x_{1}=\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}+1}{2}\right]$
+\end_inset
+
+
\end_layout
\begin_layout Subsection
in etwa der dreifachen Strichdicke.
\end_layout
+\begin_layout Standard
+\begin_inset Newpage newpage
+\end_inset
+
+
+\end_layout
+
\begin_layout Subsection
Durchgestrichene Formeln
\begin_inset Index idx
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\left(\frac{1}{1+x^{3}}\right)dx=-\frac{\log\left(x^{2}-x+1\right)}{6}+\frac{\arctan\left(\frac{2\,x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log\left(x+1\right)}{3}$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\lang english
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+\lang english
+Note:
+\series default
+ One needs to use proper delimiter insets
+\begin_inset Formula $\left(\right)$
+\end_inset
+
+ instead of simple '(' ')' characters.
+\end_layout
+
+\end_inset
+
+
\end_layout
\begin_layout Itemize
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit{i}_{2}}=0}^{\infty}{\frac{4^{-{\mathit{i}_{2}}-1}\,\left(x-1\right)^{{\mathit{i}_{2}}+1}}{{\mathit{i}_{2}}+1}}-\log4$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $solve\left(x_{1}+y_{1}^{3}=y_{1}+x_{1}^{2},x_{1}\right)=\left[x_{1}=-\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}-1}{2},x_{1}=\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}+1}{2}\right]$
+\end_inset
+
+
\end_layout
\begin_layout Subsection
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\left(\frac{1}{1+x^{3}}\right)dx=-\frac{\log\left(x^{2}-x+1\right)}{6}+\frac{\arctan\left(\frac{2\,x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log\left(x+1\right)}{3}$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\lang english
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+\lang english
+Note:
+\series default
+ One needs to use proper delimiter insets
+\begin_inset Formula $\left(\right)$
+\end_inset
+
+ instead of simple '(' ')' characters.
+\end_layout
+
+\end_inset
+
+
\end_layout
\begin_layout Itemize
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit{i}_{2}}=0}^{\infty}{\frac{4^{-{\mathit{i}_{2}}-1}\,\left(x-1\right)^{{\mathit{i}_{2}}+1}}{{\mathit{i}_{2}}+1}}-\log4$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $solve\left(x_{1}+y_{1}^{3}=y_{1}+x_{1}^{2},x_{1}\right)=\left[x_{1}=-\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}-1}{2},x_{1}=\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}+1}{2}\right]$
+\end_inset
+
+
\end_layout
\begin_layout Subsection
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $\int\left(\frac{1}{1+x^{3}}\right)dx=-\frac{\log\left(x^{2}-x+1\right)}{6}+\frac{\arctan\left(\frac{2\,x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\log\left(x+1\right)}{3}$
+\end_inset
+
+
+\begin_inset Newline newline
+\end_inset
+
+
+\lang english
+
+\begin_inset Note Greyedout
+status open
+
+\begin_layout Plain Layout
+
+\series bold
+\lang english
+Note:
+\series default
+ One needs to use proper delimiter insets
+\begin_inset Formula $\left(\right)$
+\end_inset
+
+ instead of simple '(' ')' characters.
+\end_layout
+
+\end_inset
+
+
\end_layout
\begin_layout Itemize
\end_inset
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $powerseries\left(-\log\left(5-x\right),x,1\right)=\sum_{{\mathit{i}_{2}}=0}^{\infty}{\frac{4^{-{\mathit{i}_{2}}-1}\,\left(x-1\right)^{{\mathit{i}_{2}}+1}}{{\mathit{i}_{2}}+1}}-\log4$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $solve\left(x_{1}+y_{1}^{3}=y_{1}+x_{1}^{2},x_{1}\right)=\left[x_{1}=-\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}-1}{2},x_{1}=\frac{\sqrt{4\,y_{1}^{3}-4\,y_{1}+1}+1}{2}\right]$
+\end_inset
+
+
\end_layout
\begin_layout Subsection