换元积分法 ∫x/(x^2-x-2)dx
问题描述：
换元积分法 ∫x/(x^2-x-2)dx
答
∫xdx/(x^2-x-2)=(1/3)∫[(x-2)+2*(x+1)]dx/[(x-2)(x+1)]=(1/3)∫dx/(x+1)+(2/3)∫dx/(x-2) s=x+1 t=x-2=(1/3)∫ds/s+(2/3)∫dt/t=(1/3)lns+(2/3)lnt+C=(1/3)ln|(x+1)|+(2/3) ln|(x-2)|+C