Ответ (Неопределённый)
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| asin\e / dx = C + x*asin\e / + | -------------------------- dx
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$${{2\,\int {{{x\,e^{{{\log \left(e^{x}+1\right)}\over{2}}+{{\log
\left(e^{x}-1\right)}\over{2}}+2\,x}}\over{\left(e^{2\,x}-1\right)\,
e^{\log \left(e^{x}+1\right)+\log \left(e^{x}-1\right)}+e^{2\,x}-1}}
}{\;dx}-2\,i\,\int {{{x\,e^{2\,x}}\over{\left(e^{2\,x}-1\right)\,e^{
\log \left(e^{x}+1\right)+\log \left(e^{x}-1\right)}+e^{2\,x}-1}}
}{\;dx}+i\,x\,\log \left(e^{x}+1\right)+i\,{\it li}_{2}(e^{x})+i\,
{\it li}_{2}(-e^{x})+i\,x\,\log \left(1-e^{x}\right)-i\,x^2+2\,
{\rm atan2}\left(1 , \sqrt{e^{x}-1}\,\sqrt{e^{x}+1}\right)\,x}\over{
2}}$$
1
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0
$$\int_{0}^{1}{\arcsin e^ {- x }\;dx}$$
=
1
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| asin\e / dx
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0
$$\int\limits_{0}^{1} \operatorname{asin}{\left(e^{- x} \right)}\, dx$$