MathematicsHard64×since 2004Q4023If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}∫(e2x+2ex−e−x−1)e(ex+e−x)dx = g(x)e(ex+e−x)+cg\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + cg(x)e(ex+e−x)+c where c is a constant of integration, then g(0) is equal to :A1B2CeDe²Check answerSkip