将被积函数分解为最简分式之和,
即令1/[(x+1)(x+2)^2(x+3)^3]=A/(x+1)+(Bx+C)/(x+2)^2+(Dx^2+Ex+F)/(x+3)^3
确定A,B,...,F。
可积出得:
原式=5/4/(x+3)+2*lnIx+2I+1/(x+2)-17/8*lnIx+3I+1/4/(x+3)^2+1/8*lnIx+1I+C