∫ex(sinhx+coshx)dx Let sinhx=f(x)2ex−e−x&coshx=g(x)=2ex+e−x Now, ∫ex[f(x)+g(x)]dx⇒∫exf(x)dx+∫exg(x)dx⇒∫exf(x)dx+exg(x)−∫exg′(x) ........(1) ∵g(x)=2ex+e−xg′(x)=f(x). ....(2) By eq. (1) We have, ∫exf(x)dx+exg(x)−∫exg′(x)dx By using equation (2) ⇒∫exf(x)dx+exg(x)−∫exf(x)dx+C⇒exg(x)+C⇒excoshx+C