(1)letu=√x2udu = dx∫ [cos(√x -1)/(2√x) ] dx=∫ [cos(u -1)/(2u) ] ( 2u du)=∫ cos(u -1) du=sin(u-1) + C=sin(√x-1) + C(2)∫ dx/(x^2-4)=(1/4)∫ [1/(x-2) - 1/(x+2)] dx=(1/4)ln| (x-2)/(x+2)| + C
