If ∫e2x+2ex−e−x−1e(ex+e−x)dx=g(x)e(ex+e−x)+C, where C is constant of integration then the value of g(0) is

Given ∫e2x+2ex−e−x−1e(ex+e−x)dx=g(x)e(ex+e−x)+C,Differentiate both sides ⇒e2x+2ex−e−x−1eex+e−x=g'xeex+e−x+gxeex+e−xex−e−x⇒e2x+2ex−e−x−1=g'x+gxex−e−xe2x+ex−e−x−exe-x+ex=g'x+gxex−e−x (taking 1=exe-x)ex(ex+1)-e-x(ex+1)+ex=g'x+gxex−e−xex+1)(ex−e−x+ex=g'x+gxex−e−x  Therefore, g(x)=ex+1,  it gives g(0)=2