axiom integrate(log(1-z^3)*(%i*z)^(1/2), z)
Type: Union(Expression(Complex(Integer)),...) another instance --kratt6, Tue, 13 Jan 2009 02:09:57 -0800 reply axiom integrate(sin(z)*csc(z)*(1-1/(%i*z)^(1/2))^(1/2), z)
Type: Union(Expression(Complex(Integer)),...) axiom integrate((1-1/(%i*z)^(1/2))^(1/2), z)
Type: Union(Expression(Complex(Integer)),...) |
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last edited 15 years ago by kratt6 |
![\label{eq1}{\left(
\begin{array}{@{}l}
\displaystyle
{{\sqrt{{16}\ i}}\ {\log \left({{{4 \ i \ z \ {\sqrt{i \ z}}}+{i \ {\sqrt{{16}\ i}}}}\over 4}\right)}}-
\
\
\displaystyle
{{\sqrt{{16}\ i}}\ {\log \left({{{4 \ i \ z \ {\sqrt{i \ z}}}-{i \ {\sqrt{{16}\ i}}}}\over 4}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\log \left({-{z^3}+ 1}\right)}}-{8 \ z}\right)}\ {\sqrt{i \ z}}}](images/906472983081002455-16.0px.png "\label{eq1}{\left(
\begin{array}{@{}l}
\displaystyle
{{\sqrt{{16}\ i}}\ {\log \left({{{4 \ i \ z \ {\sqrt{i \ z}}}+{i \ {\sqrt{{16}\ i}}}}\over 4}\right)}}-
\
\
\displaystyle
{{\sqrt{{16}\ i}}\ {\log \left({{{4 \ i \ z \ {\sqrt{i \ z}}}-{i \ {\sqrt{{16}\ i}}}}\over 4}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\log \left({-{z^3}+ 1}\right)}}-{8 \ z}\right)}\ {\sqrt{i \ z}}}")
![\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}+ 1}\right)}}-
\
\
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}- 1}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}}](images/4412734653952414940-16.0px.png "\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}+ 1}\right)}}-
\
\
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}- 1}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}}")
![\label{eq3}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}+ 1}\right)}}-
\
\
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}- 1}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}}](images/8161016195241591393-16.0px.png "\label{eq3}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}+ 1}\right)}}-
\
\
\displaystyle
{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}- 1}\right)}}+
\
\
\displaystyle
{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{{{\sqrt{i \ z}}- 1}\over{\sqrt{i \ z}}}}}")