作者mk426375 (时雨)
看板Math
标题Re: [微积] 积分一题
时间Thu May 19 01:20:14 2011
※ 引述《pentiumevo (神秘数学组织SIGMA)》之铭言：
: 求积分
:
: \int sqrt(x^3 + x^4) dx
= \int x*sqrt(x+x^2) dx
(Let u=x+x^2, then du=(1+2x)dx)
= \int (1/2)*sqrt(x+x^2)(1+2x) dx - 1/2 \int sqrt(x+x^2) dx
= 1\2 \int sqrt(u) du - 1/2 \int sqrt(x+x^2) dx
= 1/3*u^(3/2) - 1/2 \int sqrt[(x+1/2)^2-(1/2)^2] dx
= 1/3*(x+x^2)^(3/2) - 1/2 \int sqrt[(x+1/2)^2-(1/2)^2] dx
\int sqrt[(x+1/2)^2-(1/2)^2] dx
(Let u=x+1/2, then du=dx)
=\int sqrt(u^2-(1/2)^2) du
(Let u=(1/2)*sec(theta), then du=(1/2)*sec(theta)tan(theta)d(theta))
=\int (1/2)*tan(theta)*(1/2)*sec(theta)tan(theta) d(theta)
=1/4 \int sec^3(theta)-sec(theta) d(theta)
=1/4 [(1/2)*sec(theta)tan(theta)-(1/2)*ln|sec(theta)+tan(theta)|] + C
x+1/2=1/2*sec(theta) => sec(theta)=2x+1 => tan(theta)=sqrt(4x^2+2x)
replace the above theta in terms of x
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◆ From: 140.114.201.140
※ 编辑: mk426375 来自: 140.114.201.140 (05/19 01:22)
※ 编辑: mk426375 来自: 140.114.201.140 (05/19 01:29)
1F：推 pentiumevo :谢谢 05/19 10:00