3-x对x进行不定积分是多少啊 求大神解答下,要下详细的步骤,手写拍照的照片也可以啊。 20
1个回答
2017-11-07
展开全部
∫x(3-x^2)^3dx
=1/2∫(3-x^2)^3*2xdx
=1/2∫(3-x^2)^3dx^2
=-1/2∫(3-x^2)^3d(-x^2)
=-1/2∫(3-x^2)^3d(3-x^2)
=-1/2* 4*(3-x^2)^4+C
=-2(3-x^2)^4+C
=1/2∫(3-x^2)^3*2xdx
=1/2∫(3-x^2)^3dx^2
=-1/2∫(3-x^2)^3d(-x^2)
=-1/2∫(3-x^2)^3d(3-x^2)
=-1/2* 4*(3-x^2)^4+C
=-2(3-x^2)^4+C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询