∫(e2x+2ex−e−x−1).e(ex+e−x)dx I=∫(e2x+ex−1).e(ex+e−x)dx+∫(ex−e−x)e(ex+e−x)dx =∫ex(ex+1−e−x).e(ex+e−x)dx+e(ex+e−x) =∫(ex−e−x+1)e(ex+e−x+x)dx+e(ex+e−x) Let ex+e−x+x=t⇒(ex+e−x+1)dx=dt =∫etdt+e(ex+e−x)=et+e(ex+e−x)+C =e(ex+e−x+x)+e(ex+e−x)+C =(ex+1).e(ex+e−x)+C So, g(x)=1+ex and g(0)=2