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Let f(x)=1ex+8e−x+4e−3x and g(x)=1e3x+8ex+4e−x. If ∫(f(x)−2g(x))dx=h(x)+C, where C is constant of integration and limx→∞h(x)=π4, then the value of 2tan(2h(0)) is

Answer» Let f(x)=1ex+8ex+4e3x and g(x)=1e3x+8ex+4ex. If (f(x)2g(x))dx=h(x)+C, where C is constant of integration and limxh(x)=π4, then the value of 2tan(2h(0)) is


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