1／[x平方·根号下（2x-1）] 的积分
1个回答
展开全部
令√(2x-1)=t,则x=(t^2+1)/2
原式=∫1/((t^2+1)^2/4*t)*tdt
=4∫dt/(t^2+1)^2
由于∫dt/(t^2+1)^n
=t/(t^2+1)^n-∫t*(-n)/(t^2+1)^(n+1)*2tdt
=t/(t^2+1)^n+2n∫(t^2+1-1)/(t^2+1)^(n+1)dt
=t/(t^2+1)^n+2n∫dt/(t^2+1)^n-2n∫dt/(t^2+1)^(n+1)
所以∫dt/(t^2+1)^(n+1)=1/(2n)*(t/(t^2+1)^n+(2n-1)∫dt/(t^2+1)^n)
令n=1,代入原式得原式=4*1/2*(t/(t^2+1)+∫dt/(t^2+1))
=2t/(t^2+1)+2arctant+C
原式=∫1/((t^2+1)^2/4*t)*tdt
=4∫dt/(t^2+1)^2
由于∫dt/(t^2+1)^n
=t/(t^2+1)^n-∫t*(-n)/(t^2+1)^(n+1)*2tdt
=t/(t^2+1)^n+2n∫(t^2+1-1)/(t^2+1)^(n+1)dt
=t/(t^2+1)^n+2n∫dt/(t^2+1)^n-2n∫dt/(t^2+1)^(n+1)
所以∫dt/(t^2+1)^(n+1)=1/(2n)*(t/(t^2+1)^n+(2n-1)∫dt/(t^2+1)^n)
令n=1,代入原式得原式=4*1/2*(t/(t^2+1)+∫dt/(t^2+1))
=2t/(t^2+1)+2arctant+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
