1.If ∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c\int(e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx = g(x)e^{(e^x+e^{-x})}+c∫(e2x+2ex−e−x−1)e(ex+e−x)dx=g(x)e(ex+e−x)+c, then g(0)=g(0)=g(0)=a.eb.e^2c.1d.2Login to continueOnly logged in users canattempt or see the solution.
