changed: - \begin{axiom} f := x/(x^3-x+1) i := integrate(f, x) D(i, x) 23*%%F0^3 + %%F0 + 1 \end{axiom}
f := x/(x^3-x+1)
i := integrate(f, x)
D(i, x)
23*%%F0^3 + %%F0 + 1
![\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{
\begin{array}{@{}l}
\displaystyle
4 \ {\sqrt{{{69}\ {{\%\%F 0}^{2}}}+ 4}}\ \cdot
\
\
\displaystyle
{\arctan{\left({{{\left({9 \ {\sqrt{23}}\ \%\%F 0}+{\sqrt{23}}\right)}\ {\sqrt{{{6
9}\ {{\%\%F 0}^{2}}}+ 4}}}\over{{{10}\ {\sqrt{{{\left(-{{207}\ x}+{46}\right)}\ {{\%\%F 0}^{2}}}+{{\left({{23}\ x}-{69}\right)}\ \%\%F 0}+{{25}\ {{x}^{2}}}-{6 \ x}- 7}}}-{{207}\ {{\%\%F 0}^{2}}}+{{23}\ \%\%F 0}+{{50}\ x}- 6}}\right)}}](images/1190885338407793293-16.0px.png "\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{
\begin{array}{@{}l}
\displaystyle
4 \ {\sqrt{{{69}\ {{\%\%F 0}^{2}}}+ 4}}\ \cdot
\
\
\displaystyle
{\arctan{\left({{{\left({9 \ {\sqrt{23}}\ \%\%F 0}+{\sqrt{23}}\right)}\ {\sqrt{{{6
9}\ {{\%\%F 0}^{2}}}+ 4}}}\over{{{10}\ {\sqrt{{{\left(-{{207}\ x}+{46}\right)}\ {{\%\%F 0}^{2}}}+{{\left({{23}\ x}-{69}\right)}\ \%\%F 0}+{{25}\ {{x}^{2}}}-{6 \ x}- 7}}}-{{207}\ {{\%\%F 0}^{2}}}+{{23}\ \%\%F 0}+{{50}\ x}- 6}}\right)}}")
