If ∫ (e²ˣ + 2eˣ - e⁻ˣ - 1)e^((eˣ + e⁻ˣ))dx = g(x)e^((eˣ + e⁻ˣ)) + c where c is a constant of integration, then g(0) is equal to:

Integral Calculus: JEE Main 2020 Mathematics question

JEE Main 2020 · 5 Sep, Morning shift · Q4023

JEE Main 20205 Sep, Morning shiftMathematicsHardIntegral Calculus369× since 2002

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c

where c is a constant of integration, then g(0) is equal to :

JEE Main 2020 · 5 Sep, Morning shift · Q4023

JEE Main 20205 Sep, Morning shiftMathematicsHardIntegral Calculus369× since 2002

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c

where c is a constant of integration, then g(0) is equal to :