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1、p?(99?)j27:?*3D?單元27:積分練習與部分分式 (7.3).?*3?“複?k?,%* 3,培|?g?,v6欣?型 j巧妙5T.W1.t.? Z tanxsec2xetanxdx根W?%,lJHp,Iw = tanx dw = sec2xdx) Z tanxsec2xetanxdx =Z wewdw_?j?2型G.Iu = w du = dwdv = ewdw v = ew12b:PMp?(99?)j27:?*3D?)|?=Z wewdw=wewZ ewdw=wew ew+ C=(tanx)etanx etanx+ C=etanx(tanx 1) + CW2.tl?Z301 9 +
2、 x2dxl根W?函b型?,複3_d Z1 1 + x2dx = tan1x + CC它R? Z1 1 + (ax)2dx =1 atan1(ax) + CCy般R? Z1 1 + (k + ax)2dx =1 atan1(k + ax) + C22b:PMp?(99?)j27:?*3D?QO,b?函b?ADd2?函br ?(,n?Uvd.-:?=1 9Z301 1 + (x/3)2dx=1 9 3tan1?x3?30=1 3“tan133! tan10#=1 3?6 0?= 18W3.t.? Z x1/2ln? x1/2ex? dx:?“?l“?.根Wb函b4”, ?“?A?=Z x1/2?
3、 lnx1/2+ lnex? dx=1 2Z x1/2lnxdx +Z x1/2xdxdef=I1+ I232b:PMp?(99?)j27:?*3D?QO,?I1DI2.l, I1?函bu_? 2/?j?,Iu = lnx du =1 xdxJdv = x1/2dx v =2 3x3/27|I1=1 2?23x3/2lnx 2 3Z x1/2dx?=1 3x3/2lnx 2 9x3/2+ CI2=Z x3/2dx =2 5x5/2+ C,?=I1+ I2=1 3x3/2lnx 2 9x3/2+2 5x5/2+ C.?函bD?(Rational Functions & Partial Fract
4、ions)42b:PMp?(99?)j27:?*3D?u?函bf(x) =P(x) Q(x) .?x巧,w2P(x), Q(x)?.驟 -:(1)Jdeg(P(x) deg(Q(x), (long division)Z寫f(x) = P1(x) +R(x) Q(x)(?)w2deg(R(x) (a)deg() = deg(母),l“ ?(C),y.?,) Zx x + 2dx=Zx + 2 2 x + 2dx=Z 1|z P1(x)dx +Z2 x + 2dx=x 2ln|x + 2| + C(b) deg() deg(母),%(,)3x3 7x2+ 17x 3 x2 2x + 5= 3x 1
5、 +2 x2 2x + 5 (2)2 4(1)(5) = 16 u_?函b.?,Hp 5,5?.j母,)x2 x = x(x 1) (s_?)根W驟(3),I1 x(x 1)=A x+B x 1 jA, B:si乘x(x 1),)1 = A(x 1) + Bx = (A + B)x Ab,)A + B = 0, A = 192b:PMp?(99?)j27:?*3D?,A = 1, B = 1|(,M?,)?=Z1 xdx +Z1 x 1dx=ln|x| + ln|x 1| + C=ln?x 1 x?+ CW6.t.? Z1 x2(x + 1)dx母j_?x2(x複s)D(x + 1)根W?驟(
6、3),I1 x2(x + 1)=A x+B x2+C x + 1jA, B, C:si乘x2(x + 1),)1=Ax(x + 1) + B(x + 1) + Cx2 =(A + C)x2+ (A + B)x + B102b:PMp?(99?)j27:?*3D?b,)A + C = 0, A + B = 0, B = 1,A = 1, B = 1, C = 1|(,M?,)?=Z1 xdx +Z1 x2dx +Z1 x + 1dx=ln|x| 1 x+ ln|x + 1| + CW7.t.?Z2x3 x2+ 2x 2 (x2+ 2)(x2+ 1)dx母j_.j?x2+ 2Dx2+ 1I2x3
7、x2+ 2x 2 (x2+ 2)(x2+ 1)=Ax + B x2+ 2+Cx + D x2+ 1112b:PMp?(99?)j27:?*3D?jA, B, C, D:si乘(x2+ 2)(x2+ 1),)2x3 x2+ 2x 2=(Ax + B)(x2+ 1) + (Cx + D)(x2+ 2)=(A + C)x3+ (B + D)x2+ (A + 2C)x +(B + 2D)b,)A + C=2(1)B + D=1(2)A + 2C=2(3)B + 2D=2(4)QO,(3) (1) :C = 0,(1), A = 2 (4) (2) :D = 1,(2), B = 0,?=Z2x x2+
8、 2dx +Z1 x2+ 1dx=Z1 udu tan1x(u = x2+ 2, du = 2xdx)=ln|u| tan1x + C=ln(x2+ 2) tan1x + C122b:PMp?(99?)j27:?*3D?W8.t.?Zx2+ x + 1 (x2+ 1)2dx母_.j?(x2+ 1)2(x2+ 1複s)Ix2+ x + 1 (x2+ 1)2=Ax + B x2+ 1+Cx + D (x2+ 1)2jA, B, C, D:si乘(x2+ 1)2,)x2+ x + 1= (Ax + B)(x2+ 1) + Cx + D= Ax3+ Bx2+ (A + C)x + (B + D)b,)
9、A = 0, B = 1, A + C = 1, B + D = 1j5,)A = 0, B = 1, C = 1, D = 0,M?,)132b:PMp?(99?)j27:?*3D?=Z1 x2+ 1dx +Zx (x2+ 1)2dx=tan1x +1 2Z1 u2du(u = x2+ 1, du = 2xdx)=tan1x 1 21 u+ C=tan1x 1 2(x2+ 1)+ CW9.t.? Z1 (x + 1)2(x2+ 1)dx母_?(x + 1)2(x + 1複s)D_.j?x2+ 1I1 (x + 1)2(x2+ 1)=A x + 1+B (x + 1)2+Cx + D x2+
10、1142b:PMp?(99?)j27:?*3D?jA, B, C, D:si乘(x + 1)2(x2+ 1),)1=A(x + 1)(x2+ 1) + B(x2+ 1) +(Cx + D)(x + 1)2=A(x3+ x2+ x + 1) + B(x2+ 1) +(Cx + D)(x2+ 2x + 1)=(A + C)x3+ (A + B + 2C + D)x2+(A + C + 2D)x + (A + B + D)b,)A + C = 0(5)A + B + 2C + D = 0(6)A + C + 2D = 0(7)A + B + D = 1(8)QOH(5)p(7) : D = 0H(8)p(6) : 1 + 2C = 0 C = 12(5) : A 1 2= 0 A =1 2(8) :1 2+ B + 0 = 1 B =1 2152b:PMp?(99?)j27:?*3D?|(,M?,)?=Z1 2 x + 1dx +Z1 2 (x + 1)2dx +Z12x x2+ 1dx=1 2ln|x + 1| 1 2(x + 1)1 4ln(x2+ 1) + C162b:PM
