`inte^x/(1-tan(e^x))dx`
First, let's apply integral substitution:`u=e^x`
`=>du=e^xdx`
`=int1/(1-tan(u))du`
Now from the integration tables:
`int1/(1+-tan(u))du=1/2(u+-ln|cos(u)+-sin(u)|)+C`
Using the above,
`=1/2(u-ln|cos(u)-sin(u)|)+C`
Substitute back `u=e^x`
`=1/2(e^x-ln|cos(e^x)-sin(e^x)|)+C`
`inte^x/(1-tan(e^x))dx`
First, let's apply integral substitution:`u=e^x`
`=>du=e^xdx`
`=int1/(1-tan(u))du`
Now from the integration tables:
`int1/(1+-tan(u))du=1/2(u+-ln|cos(u)+-sin(u)|)+C`
Using the above,
`=1/2(u-ln|cos(u)-sin(u)|)+C`
Substitute back `u=e^x`
`=1/2(e^x-ln|cos(e^x)-sin(e^x)|)+C`
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