求∫x^3÷√(1-x^2)dx的不定积分

求∫x^3÷√(1-x^2)dx的不定积分 求I=?x^3/?(1-x^2)dx的不定积分 解法一: 思路:根据分子分母的关系，直接变形化简求得: I=-?[x(1-x^2)-x]dx/?(1-x^2) =-?x(1-x^2)dx/?(1-x^2)+ ?xdx/?(1-x^2) =-?x?(1-x^2)dx-(1/2) ?d(1-x^2)/?(1-x^2) =(1/2) ??(1-x^2)d(1-x^2)-?(1-x^2) =(1/3)?(1-x^2)^3-?(1-x^2)+c 解法二: 思路:利用不定积分的分部积分方法求得: I=?x^2*xdx/?(1-x^2) =-(1/2)?x^2d(1-x^2)/?(1-x^2) =-?x^2d?(1-x^2) =-x^2?(1-x^2)+ ??(1-x^2)dx^2 =-x^2?(1-x^2)-??(1-x^2)d(1-x^2) =-x^2?(1-x^2)-(2/3)?(1-x^2)^3+c 解法三: 思路:利用三角函数的代换关系求得。 设x=sint，则cost=?(1-x^2),此时: I=?sin^3td(sint)/?(1-sin^2t) =?sin^3t*cost dt/cost =?sin^3 t dt =?sint(1-cos^2 t)dt =?sintdt-?sintcos^2 tdt =-cost+?cos^2tdcost =-cost+(1/3)cos^3 t+c =-?(1-x^2)+ (1/3)?(1-x^2)^3+c.