4个回答
展开全部
换一种思路:由微分求积分
[(1/3)x^3 arctan(3x)]'
= x^2 arctan(3x) + x^3/(1+9x^2)
= x^2 arctan(3x) + (1/9)x - (x/9)/(1+9x^2)
Therefore,
∫x^2 arctan(3x) dx = (1/3)x^3 arctan(3x) - x^2/18 + (1/162)ln(1+9x^2) + c
[(1/3)x^3 arctan(3x)]'
= x^2 arctan(3x) + x^3/(1+9x^2)
= x^2 arctan(3x) + (1/9)x - (x/9)/(1+9x^2)
Therefore,
∫x^2 arctan(3x) dx = (1/3)x^3 arctan(3x) - x^2/18 + (1/162)ln(1+9x^2) + c
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
∫x^2 arctan3x dx = (1/3)∫arctan3x dx^3
= (1/3) x^3 arctan3x - (1/3)∫[3x^3/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/9)∫[(9x^3+x-x)/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/9)∫[x-x/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/18)x^2 + (1/9)∫[x/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/18)x^2 + (1/162)ln(1+9x^2) + C
= (1/3) x^3 arctan3x - (1/3)∫[3x^3/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/9)∫[(9x^3+x-x)/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/9)∫[x-x/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/18)x^2 + (1/9)∫[x/(1+9x^2)] dx
= (1/3) x^3 arctan3x - (1/18)x^2 + (1/162)ln(1+9x^2) + C
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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