计算积分:∫√[(2+3x)/(x-3)] dx
2个回答
展开全部
令√[(2+3x)/(x-3)]=t,则x=(3t²+2)/(t²-3)
∫√[(2+3x)/(x-3)]dx
=∫td[(3t²+2)/(t²-3)]
=(3t²+2)t/(t²-3) -∫[(3t²+2)/(t²-3)]dt
=(3t²+2)t/(t²-3)- ∫[3+ 11/(t²-3)]dt
=(3t²+2)t/(t²-3)-3t -[11/(2√3)]∫[1/(t-√3)-1/(t+√3)]dt
=(3t²+2)t/(t²-3)-3t -(11√3/6)ln|(t-√3)/(t+√3)| +C
=11t/(t²-3) -(11√3/6)ln|(t²-2√3t+3)/(t²-3)| +C
=11√[(2+3x)/(x-3)]/((2+3x)/(x-3) -3]
-(11√3/6)ln|[(2+3x)/(x-3)-2√[(6+9x)/(x-3)] +3)/[(2+3x)/(x-3)]-3]| +C
∫√[(2+3x)/(x-3)]dx
=∫td[(3t²+2)/(t²-3)]
=(3t²+2)t/(t²-3) -∫[(3t²+2)/(t²-3)]dt
=(3t²+2)t/(t²-3)- ∫[3+ 11/(t²-3)]dt
=(3t²+2)t/(t²-3)-3t -[11/(2√3)]∫[1/(t-√3)-1/(t+√3)]dt
=(3t²+2)t/(t²-3)-3t -(11√3/6)ln|(t-√3)/(t+√3)| +C
=11t/(t²-3) -(11√3/6)ln|(t²-2√3t+3)/(t²-3)| +C
=11√[(2+3x)/(x-3)]/((2+3x)/(x-3) -3]
-(11√3/6)ln|[(2+3x)/(x-3)-2√[(6+9x)/(x-3)] +3)/[(2+3x)/(x-3)]-3]| +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