Front | \[\int e^{ax} \tanh bx\ dx \] |
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Back | \[\begin{cases}\displaystyle{ \frac{ e^{(a+2b)x}}{(a+2b)}{_2F_1}\left[ 1+\frac{a}{2b},1,2+\frac{a}{2b}, -e^{2bx}\right] }& \\\displaystyle{\hspace{1cm}-\frac{1}{a}e^{ax}{_2F_1}\left[ 1, \frac{a}{2b},1+\frac{a}{2b}, -e^{2bx}\right]}& a\ne b \\ \displaystyle{\frac{e^{ax}-2\tan^{-1}[e^{ax}]}{a} } & a = b\end{cases}\] |
Tags: hyperbolic
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