1、复杂因式【不定积分的第一类换元法】已知 J = F(u) + C求 J g(x)dx = J f (p(x)(px)dx = | 【凑微分】= F(“) + C 【做变换,令u =(px)再积分】=F(x) + C 变量还原,=(p(x)【求不定积分gMdx的第一换元法的具体步骤如下:1(1)变换被积函数的积分形式:,(兀)心=7(傾功0(.巧心 凑微分:Jg(x)x=J/(0(x)0a)x=“(0(x)心心)(3)作变量代换u =(p(x)得:Jg(x)& = J/0X)0G皿=打(傾功处:)=j f(u)du(4)利用基本积分公式J/(“)血=F(W) + C求岀原函数:J gWdx =
2、| f(p(x)(px)dx = J /(仅 x)傾 x) = j f(u)du = F(u) + C(5)将=(p(x)代入上面的结果,回到原来的积分变量x得:J gWdx = J/(0(x)0G)dx = j f(p(x)d(px) = = F(u) + C = F(p(x) + C【注】熟悉上述步骤后,也可以不引入中间变u =(p(x).省略步骤,这与复合函数 的求导法则类似。二、典型例题 J f (ax + b)dx =丄 J f (ax + b)dax + Z?f w I-Jl + x2 + Jq + x解:令“ =2x-1, du = 2tZv,J角弓2tdt _ 1 f (f +
3、 1 - l)d= = ?JrT7= L|(r + 1)Ll.2Vr77+c=l(+i)L+c打 (1)+打上-4J肛孑 2肛卩令 l + F =t_ f dt + 1) =J VCT= 2jl + VF +C = 2jl + Jl + / +c=-丄 x2x Jl-x4 + arcsinx2 + C4 2=(arcsinx2 _ l-x4 ) + C2(2)j/(suix)cosay/x = J/(sinx)6/ sin 小 J/(cosx)siii xdx = -J/(cosx)cosx,f /(tan牙) =f/(tanx)Jtanx , f /(cotx) = - f f(cotxl
4、cotx ;J cos x J J sin x J + + lnlcscx - cotxl + C3cos x cosxdx5解:f-sin xcos7 x=tanxcos4 xsin2 x + cos2 xtanxcos2 xJ tan xsin u f ,1 , cos + sin* u dv2J 1 + v2e tan xd tan xa1 tan2 x + Z?f 1 + tan x . 1 - .I= a tanx = tan x + In tan.v + CJ tnn r 2有sin2x - 1 r j dx = -Jcos 2x + sin2 2x1e J COSH _ 1 r=
5、F ii =_RCOS M + COS* It2 2=-arctanv + C = - arctan(cos2x) + C22r s r tanx f7-解:hda宀+胃3J /(In x)丄 dx = J /(In x)d In x, J f(ex )exdx = J f(e x )des ;42.解令 = 5x,弓忤T吋cdu = 5dx5J e5xdx = 1 f edu = -eu +C = -e5x +C3解:令u=3 + 4e du = 4eKdx9= arcsin(bix) + C十亠 + cx X(1+)2 1 +即32=2xylex - 2 - 2J yjex - 2(lxe
6、x -2 = r2, ex =2 + r2, x = ln(2 + r2), dx = , dt2 +广原式=2応F 一口 寻妇2皿二-4J盲二(丄)(-)J 兀 X。 J X X=2x/ex -2 -4r + 8 -= arctan -L- + CV2 yl2=- 2 _ 4 Je* - 2 + 4、5 iirctun i 1- C8解:f n 人 dx = f 八 d tanx = f lii tan x(ln tan x) J cosxsinx tanx J(In tanx)2 厂6f(xn)xndx = -ff(xn)dxn (心 0),J = 2J f(yx)d(x);d x例2.
7、J例倚例6. f,办 (a 0) J yjx(a-x)1.解:dyx =丄上22Vx3丘 _/x vdx = - f (Vl + x2 - 1 /(l + x2)2 J17?希j晋心晋心芒h諾对于右端第一个积分,凑微分得.dx = -(1 -x2) 26/(1-x2) = -Vl-x2 +CJ Vl-x2 j第二个积分中,用代换x = sintr X2 f r sin / , f 1 - cos21 , dx = cos tdt = ItJ 71-X2 cosr J 2=- -sin 2t + C= arcsinx - xVl - x2 + C2 4 2 2原式=1arcsinx-丄(x +
8、2)J1-F +C4 解:J ; = JJ x( Jin x + Jin x + b) J=! f y/nx + ad(n x + a) + ! f y/nx + bd(n x + b) a -b) a-bJ=-2jl-x arcs in 4x + 2yx + C7打(arcs in x) /2 j = I2 Intan 6解:J 1 +COSX JQ(1 + sinx)(l -cosx)氐1 -cos x訂-J sm xe cosx f ;UX + sin - xdx 一 f ex cot xdx sinx =exd( !) +J J sinxdx - ex cot xdxsinx J=-ex cot x + b Csinx1任下列各式等号右端的空白处填入适当的系数,使等式成立:-1).Av =d(ov+b) (“HO):xdv=d(5x2):x5dx=d(3 宀 2);e 2 dv=d(l+e 2);= d(l-arcsinv):-X2(11)cLvc )= d (arctan3x);1+9.L
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