第六章不定积分习题参考答案
第六章不定积分习题解答
练习 6.1
1. 若F(x)=f(x),则F(x)是f(x)的原函数,f(x)的原函数全体称为f(x)
的不定积分。
区别是:f(x)的不定积分描述了所有满足导数是f(x)的函数,而原函数只是任一个满足导数是f(x)的函数。 2. (1)ex3
(2)cosxc (3)1a
(4)2 (5)-1
(6)12
(7)1
2
3.(1)(10xc)10 10dx10xc (2)(2cosxc)sinx 2sinxdx2cosxc (3)d(x5c)5x4dx 5x4dxx5c 3
4.
解:由题意f(x)23
x2c,又由 f(1)1,知 c13,
f(x)23因此 1
3x23
。
5.解:由题意f(x)(lnx)1
,所以 f(x)1xx2
x21
1.(1)原式lnxc
2x
(2)原式arcsinx3cosx 2xc (3) 原式=
1e1x
xeeexc e1
4 (4)原式=(4x26x9x)dx
1x2x1x
469c ln4ln6ln9
8 (5)原式=(xxx)dxxdxxc
15
45
(6)原式=xdxxc
5
=
x4111x32
dx(x12)dxxarctanxc (7)原式=2
x1x13
x2x21111
(8)原式=2222arctanxc
x(x1)x1xxx1cosxxsinx
c
222211
(10)原式=(22)dxtanxcotxc
sinxcosx
(9)原式=cos2dx
(11)原式=(csc2x1)dxcotxxc 2. 解:由题意知C(x)C(x)dx7xxc
由固定成本为 1000 知 c=1000 因此
C(x)7x1000
8
(1)(5x3)
dx
1189
(5x3)d(5x3)(5x3)c545
3
11
(2)(2x21)(2x21)2c
46
(3)3
2
(1lnx)(1lnx)2c
3
(4)exdxdexexc
csc2xcscxcotxd(cscxcotx)
(5)cscxdxdxln|cscxcotx|c
cscxcotxcscxcotx
(6)sin2xcos4xdx
1222
4sinxcosxcosxdx4
11cos2x1sin22xdx(sin22xsin22xcos2x)dx428
11cos4x12
dxsin2xdsin2x8216111
(xsin4x)sin32xc16448
x2x2111(7)dxdx(x1)dx
dxx2xln|1x|c
x1x1x1
(8)
(9)
(10)
x11122dxln(x1)cc x212x212
111
(x1)c (x1)2
x22x1x21
1111111x1
dxlnc 2x2x3(x3)(x1)4x14x34x3
(11)
111x1dx(x1)arctanc (x1)222
x22x522
11
f(axb)d(axb)(axb)c aa
2.用第二换元积分法求下列不定积分
(12)f(axb)dx
t2t2
(1)dt2dt2t2lntc2ln(1c
1t1t
t113t2324
(2)(1x)dt(ttln|1t|)c
441t4
2
3343
(1x)ln|1c44
3
(4) 解:令xt dx2dt2
sectcost1
dtc223tant3sint3sint
原式
c
11
(5)3xln3xc
33(6)令xasint dx=acostdt a2sin2tacostdtdx= =a2sin2tdt
acost1cos2ta21
adt(tsin2t)c
222
a2a2
tsintcostc22a2x1
arcsinc2a2
(x2)lnx2c2
2
(7)lnx2c
练习6.4
1. (1) xlnxdx
112122
lnxdxxlnxxdlnx 222
12112x2
xlnxxdxxlnxc 2224
(2) x2sinxdxx2dcosxx2cosx2xcosxdx
x2cosx2xdsinxx2cosx2xsinx2sinxdxxcosx2xsinx2cosxc
xcosx(1)1x1
(3)3xdcsc2xcsc2xcsc2xdx
sinx222
2
x1
csc2xcotxc 22
xcosx(2)1x122
xdcotxcotxcot2xdx sin3x222
x1x1x
cot2x(csc2x1)dxcot2xcotxc 22222
x21x2x11
(4) xd22222(1x)2(1x)21x
1x11x1
arctanxc 21x221x222x22
(5)sec3xdxsecxdtanxsecxtanxtanxdsecx
secxtanxsecxtan2xdxsecxtanxsecx(sec2x1)dxsecxtanxsecxdxsecxdxsecxtanxsec3xdxlnsecxtanx
3
因此 (6)
3secxdx
1
(secxtanxlnsecxtanx)c 2
x2a2x2
dx
设xatant
a2tan2tasect
asec2tdt
a2tan2tsectdta2tantdsect
a2(tantsectsectdtant)a2(tantsectsec3tdt)
a2
[secttantln(secttant)]
c1 2
a2x2aaxa2)c1lnxc a2
(7
)arcsinxdxxarcsinx
xarcsinx
12
(1x)2
xarcsinxc
(8)excosxdxcosxdexexcosxexdcosx
excosxexsinxdxexcosxsinxdexecosxesinxecosxdx
x
x
x
因此 excosxdxex(cosxsinx)c (9)
设
t2
costdt2tcostdt 12
2tdsint2tsint2sintdt2tsint2costcc
arctanex设euarctanuarctanu1
dxdlnuarctanud() (10) uu2exu
x
11111
arctanuarctanuarctanudu
uuuu1u211u11arctanu()duarctanulnuln(1u2)c 2
uu1uu211
xarctanexxln(1e2x)c
e2
2.解: 由已知得
sinxsinx
c或f(x)() xx
sinxsinx
xfxdxxdfxxfxfxdxx()c xx
xcosxsinxsinxsinxxccosx2c 2
xxx
f(x)dx
习题六
1.单选题
(1) C (2)A (3)C (4) D (5)B (6)C (7)B (8)B (9)A ①f(x)ex,
f(lnx)1
dxf(lnx)dlnxelnxcc xx
1
x
f(lnx)11
dx2dxc xxx
②f(x)ex,f(lnx)elnx,
(10)D(11)C (12)A (13)C (14)D
xexexex
dxdc (15)B 2
(x1)x1x1
(16)C (17)B (18)C (19)C (20)D 2.已知曲线yf(x)的切线斜率为xex,且通过原点,求曲线方程。 解:已知 f(x)xex
f(x)xexdxxdexxexexdxxexexc
由曲线过(0,0)知f(0)0.即c=1,于是 f(x)xexex1
3.设边际收益为f(x)202x,其中x为产量,且当产量为零时,收益为零,求收益函数。
解:由边际收益函数为 f(x)202x
0x2d)x知收益函数 R(x)(2
2
2x0x C
又R(0)0 知C=0,于是 R(x)20 x2x
4.某商品的边际需求函数为Q(p)5,最大需求量为100,生产这种产品的边际成本函数为C(Q)150.05Q,固定成本为12.5元,其中
P为价格,Q为需求量,C为总成本,若供销量相等,求价格P定为
多少时,利润最大。
c100 解: Q(p)(5)dp5pc Q(0)100,
需求函数为Q(p)1005p
成本为C(Q)(150.05Q)dQ15Q0.025Q2c
2
C(0)C012.5,c12.5 C(Q)15Q0.02Q5
12.5
①利润关于P的函数为:
L(P)R(P)C(P)(1005P)P15(1005P)0.025(1005P)212.5
令L(P)0得 P
120120
L()0
77
即当P为120/7时利润最大
②L(Q)R(Q)C(Q)(20Q)Q15Q0.025Q212.5 令L(Q)0得Q
100100120
L()0 此时P
777
15
即当P为120/7时利润最大 5.函数f(x)的弹性函数为解:由已知得
1
,且f(1)0,求f(x) f(x)
Efx11f(x)即f(x)
xExf(x)f(x)
1
f(x)ln|x|c 又f(1)0 c0 得f(x)lnx
x
6.求下列不定积分
1d(3x2)1
(1) (3x2)3c 1
23
(3x2)311
(2) xsinx2dxsinx2dx2cosx2c
22
2
(3)
sinxx
dx
dcosxcosx
2cosxc
(4) excosexdxcosexdexsinexc
(5)
1d(52x2)12x2c
4252x252x2
x
x21d(x24x5)12ln|x24x5|c (6) 2
x4x52x4x52
(7) sin23xdx(8) (9)
1cos6xxsin6x
dxc 2212
1dlnxdlnlnxdxxlnxln(lnx)lnxlnlnxlnlnxln[ln(lnx)]c
dx
lnx1cln(x1c
(10) ecosxsinxdxecosxdcosxecosxc (11) sin5xdxsin4xsinxdx(1cos2x)2dcosx
2cos5x3
(12cosxcosx)dcosxcosxcosxc
35
2
4
(12) (13)
x
13lnx
dx
2
dx
dlnx3lnx
2
arcsin
lnxc
(1arcsinx)
2
x2
d(1arcsinx)1c
(1arcsinx)21arcsinx
(14) sin2xcos2xdx(15)
11112
sin2xdx(1cos4x)dxxsin4xc
48832
1313
arcsinxc 34
(16)
1d(1x2)
dxarcsinxx2arcsinxc
2x2x2x1
dxdex
(17) xx2xarctanexc
eee1
(18) tan4xdxtan2x(sec2x1)dxtan2xdtanxtan2xdx
tan3x
tanxdtanx(secx1)dxtanxxc
3
2
2
(19)
1422
sin4xdxcscxdxcscxdcotx(1cotx)dcotx
cot3x
cotxc
3
(20) sin3xcos5xdx
(1cos2x)cos5xdcosxcos8xcos6xc
(21)解:令 t
x3t31,x2dxt2dt
t8t5
xxdx(t1)ttdtc
85
32
2
3
2
2
1816
x
85c
85
(22)t122
dln(t1)tt21
(
11t1)dtlnclnc
t1t1t1t
t36t85
(23)6tdtdt221t1t(t4)211(t41)(t41)166()2221t1t1t
(t21)(t21)(t41)16(dt1t2)1t2
1
6(
(t6t4t21)dt2)
t1
66t1t7t52t36t3ln||c
75t1
67/665/61/21/6
xx2x6x3ln||c
75
(24)
113secttant
dt
costdtsintcc 29sect3tant99x3sect
2sec2t11(25)dxdtcostdtsintcc 338sect44(4x2)2
1
x2tant
(26)xuex
utant
sec2tdttantsect
csctdtln|csctcott|c
1ln|cln||c
u(27)(2x1)e3xxdx
1113x23x23x
(2xx)de(2xx)ee(4x1)dx 333
14x13x13x11
eed(4x1) (2x2x)e3x(4x1)de3x(2x2x)e3x
39939
14x13x43x(2x2x)e3xeec 3927
(28)exlnxdxexxdxxexexc
(29)excosxdxexdsinxexsinxsinxdex
exsinxexsinxdxexsinxexdcosx
esinx(ecosxcosxde)esinxecosxecosxdx
x
x
x
x
x
x
因此 excosxdx(sinxcosx)exc
(lnx)2lnxtt2t
(30) 解①22tdtt2etdtt2dett2etetdt2
xe
t2et2tetdtt2et2tet2etdt
2t
t
t
1
2
12
te2te2ec((lnx)2lnx2)c
x
(lnx)2(lnx)21(lnx)21212
(lnx)2lnx2dx 解②2(lnx)d
xxxxxx(lnx)21(lnx)2lnx1(lnx)2lnx2
2lnxd2(dlnx)2c
xxxxxxxx
(31)lnxdxxlnxxdlnxxlnx2lnxdx
xlnx2xlnx2xc
2
2
2
2
2
(32)et3t2dt3t2det3t2et3
etdt2
t3t2et6tdet3t2et6tet6etc6c
x2tx1x21t1
(33)①2x2dt 22x4x13t9x29
t111t2
lnt9arctanct292t2933
11x2lnx24x13arctanc233
②
x112x42
dx 22x4x132x4x13
11x21d(x24x13)d(x2)2
lnx4x13arctanc 222
2332x4x13(x2)3
x5x48x2x82
(xx13)dx *(34)3
xxxx
x(x1)8
)dx22
x(x1)x(x1)1x1
(x2x1)dx8(2)dx
x1x1x
1312d(x21)xxxlnx1428lnx32x1(x2x1
1312
xxxlnx14lnx218lnxc 3211
x3x2x3lnx14lnx18lnxc 32
2x2511
(*(35)解①4
x22x23)dx x5x26
dxdx
c
2x252x252解
②42
22
x5x6(x3)(x2)2 右边通分后再比较分子,由对应系数相等有:
ABCD0解得
AB
2
2A2B3C3D0CD5
原式
c
x23x2x23x2
*(36)2
x(x2x1)x(x1)2
x23x2ABC解① 令 22
x(x1)xx1(x1)
A(x1)2B(x1)xCx(AB)x2(2ABC)xAx23x2
则
x(x1)2x(x1)2x(x1)2
AB1
于是 2ABC3 得 A2,B1,C 6
A2
216x23x2x23x2
(dxxx1(x1)2)dx x(x1)2
x(x22x1)6x26
2lnxlnx1clnc
x1xx1
x23x2x11
解②32x22x1x22x1x(x1)2dx x(x22x1)
x11x2
3dx2[x22x1x(x1)2]dx x22x1
x11x232dx2x22x1xx22x1 x22x1
x11
72x22x1xdxx22x1
12x212622lnx
2x2x1x2x11d(x22x1)d(x1)
262lnx
2x2x1(x1)216lnx22x12lnxc
2x116x2622ln(x1)lnxclnc
2x1x1x1
(37)(38)
1x111111()dxlnc x2(x21)x21x2
x4x22x1x1
dx
sinxcosx
x2u1u22
,cosx,dxdu(万能置换)解① 令utan,sinx
21u21u21u2
2
du21dudx2sinxcosx2u1u2
u22u1
1u21u2
2c
cc
解②
sinxcosxsinxsin(x)2sin
2
x
2
xcos
x
2
x
2sin(x)cosx
)
444
11
dxcsc(x)d(x)
sinxcosx44x)
4
csc(x)cot(x)c
44*
(39)12(a0)解:令
tx(t2b),dxtdt
aa
122tmm2m
tdt
(1)dt
tmaatmatm22
(tmlntm)cmlnmc aa11x11x1x11x2
lndxlnln(ln)c (40)
1x21x21x1x41xlncosxsinx
dxlncosxdtanxtanxlncosxtanx (41)cos2xcosx
tanxlncosx(sec2x1)dxtanxlncosxtanxxc
1x2dx2设xt1tdt
(42)解①33
1t322221x1x
2
x3
u
22
1u1du1 32u
u2112duucc
uu解②令xtant dx
sec2tdt sect
tan3tsin3t1cos2t2
sectdt
dcost 3222sectcostcost1x
(
111)dcosttcarctanxc 2
costcost
4
x3
令tx
1x41t1t114
(43) dxdx 2242424441t1t1x1x
x7
1111111
lntc241t41t441t
111
ln1x4c441x4
(44)
2x1dx
2
4
1d2x112x1
c 222x1442
exex
dx()c (45) 解①2
1x1x1x
1xex1x
xeddxe解② 21x1x1x1x
x
xex
xex
xex1xex1xxx
(exe)dxe(1x)dx 1x1x1x1x
xexxexexxx
edxecc 1x1x1x
(46) 解① 令t,则
x,dx2dt
1
x1t1t
1(2)dt
t
c1 1d(t)
lnt
111
c1ln2x2
c
设t=
x
1
2
解②
u
设t
2u
2 21
sinucoscosusin66
csc(u)d(u)lncsc(u)cot(u)
c 6666sin(u)6
1
1cos(u)clnc sin(u)
6ln
c回代c
(47)
e2x
1e
x
设ext
uu414u1 u4u74u34x4x4(uu)duce11ec
7373
62
(48)
2
(x)
3
2
令
x3
5令tsecu
3
55
tanusecutanudu33
22
tanusecudu(secu1)secudu33
(secusecu)dusecudusecudu)
11(secutanulnsecutanulnsecutanu)c1
922
11
secutanulnsecutanu)
c1 22tc2
13x2c
3x26(49) 解
①1
d(x)
1
设xt
2
arcsin2tcarcsin2x1c
解
②(50)
2c
3x12
3x1
x1
2
16
x1
2
16
x1
x1216d32dx1
2x12162x116
31x1lnx22x17arctanc
224
(51)
1
4
d4x1设
4x1t
1
4
11
lntcln4x1
c 44
(52)
1
3
d3x11
ln3x13
1
cln3x1
c
3
(53)
2
1d
9x41
3x
9
3
1
ln3xc 3
7. 求下列不定积分 (1)faxbdx
11
dfaxbfaxbc aa
(2)xfxdxxdfxxfxfxdxxfxfxc (3)fx(4)
a
fxdxfxdfx
a
1
fxa1c a1
fxdfxlnfxc fxfx8.求满足f(x)xex,且f(0)0的函数f(x).
解:fxxexdxxdexxexexdxxexexc
f(0)0代入c11
(x0),求
f(x). x
1即fx fxxdxc fxxee1
x
x
9. 设f(x2)
解:令
ux2xf
u
10. 设f(sinx)cos2x,求f(x).
解:令usinxfu1u
2
x2
即fx1xfx1xdxxc
3
2
2
1
f(x)
11. 已知2exC,求f(x).
x
1
1fx1
解:(1)2fxdec fxex
xx
1
x
1
fxfxe
(2)2dxec(e)22得fxex
xxx
求12. 设fx的一个原函数是
sinx1,
2
解:1
(2x21)
41
fd(sinx2+1是f(x)的一个原函数
) 211
21c1sin(2x21)c 22f(x)12
[f(x)]c,且f(0)0,求fx 13. 若2
1x2
f(x)11x由已知111x2
解:(1).f(x)dln.f(x)c f(x)l
1x221x221x
f(x)1
dx[f(x)]2c 2
1x2
11f(x)2
有([f(x)]2)2f(x)f(x)f(x) 22
241x1x(2).由已知
f(x)
21x1x
lnc而f(0)0得c0f(x)ln
1x21x1x
*14.求下列不定积分
cos2x2cos2x1
sin2x
1sinxcosx2sin2x2sin2x
1
(sin2x2)ln2sin2xc1lnsinxcosxc
2sin2x
11x22xx22xx22x2
c (2)(x1)edxed(x2x)e
22
(1)
dxsec2x (3)2
sinx2cos2xtan2x2
tanxc sinxsinx(1sinx)sinxsin2x
dxdx (4)解①222
1sinxcosxcosxcosx112dcosx(secx1)dxtanxxcsecxtanxxc cos2xcosxsinxsinx1111
(1)dxxdx 解②
1sinx1sinx1sinx1sinx
1
1sinxdx
令utan
x2
12du
du22u1u2(u1)2
1
1u2
2222sinxcccc
x1cosxu11cosxsinxtan112sinx
sinx2sinxxc
1sinx1cosxsinx
(5
)(x23xexdx
(x2x2xexdx
[(x2x)ex(2x1)ex
3
22x2
(xx)e[(xx)e]c
32
x
(6
)
[ln(x5]d[ln(x5]
3
2[ln(x5]2c 3
12
(7)(8
)
arctan
1
arctan1darctan11(arctan1)2c
1x2xx2x
(ba)
1
b2a2
sin2x
sin2x
[(b2a2)sin2xa2]
c
*15. 求下列不定积分 (1
)a0)解:令
xatantdxasec2tdt
asec2tdt1cost112sint 2222
2atantasectasintasint112cc 2
asintax
(2
)解:令
xsectdxsecttantdt
sect1
secttantdt(1cost)dt
sec2ttant
tsintcarcsecxc
(3
)x
1122
(x(x)(4x2(4x2)
22
353
1142222222
[(4x)d(4x)(4x)(4x)c 253
(4
)
(a0)解:令xatant dx
asec2tdt 4
x
asect1cost2
dxasectdt 44424xatantasint
11(ax)11sintcc
a2sin4t3a2sin3t3a2x3
2
3
22
(5
)xxx
x12x
212xarcsinexc
22au24au
(6
)(a0)解:令ux,dxdu
1u2(1u2)2
2au24auu42
1u2u(1u2)2du8a(1u2)3
u41u3du21u3d(u21)131
解① ud23232322(1u)2(1u)2(1u)4(1u)
分部积分
1u31u313u23
[] 4(1u2)2(1u2)24(1u2)24(1u2)2
u3311[]du 222224(1u)41u(1u)
u331u[arctanu(arctanu)]c1(由
P30递推公式I2)
4(1u2)2421u2u333uarctanuc1
222
4(1u)881u
33
c1
28814(1)2ax2a
x3c1 8
3arctanc8原式8a2(3a2c
AuA2uB2A3uB3u41B1
解② 2322223
(1u)1u(1u)(1u)
5AuB1u4(2A1A2)u3(2B1B2)u2(A1A2A3)uB1B2B31
(1u2)3
两边分子对应系数相等有
A10,B11,A20,B22,2A1A20,2B1B20, A1A2A30,B1B2B30,A30,B31
u4121
[(1u2)31u2(1u2)2(1u2)3]du(由P30递推公式I1,I2,I3)
1uu3u3
arctanu2(arctanu)arctanuc1 2222
21u4(1u)8(1u)8
u5u3arctanuc1
4(1u2)28(1
u2)8
3c1 8
3c
8原式8a2(*16.
3a2arctanc
求下列不定积分
xarctanx11arctanxd
(1x2)221x2
(1)
1arctanx111arctanx1dx
darctanx(据P30I2) 2222221x21x21x2(1x)
1arctanx1xx21x
arctanx)carctanxc
21x24x21(4x21)(4x21)
(2)
x1)
(x1)
d(x11
1xx
dx2
1x)
(x1)arcsin
(x1)(
x1)(x1)arcsinc
arcsinx2(3)
x2sgnxarcsinxdarcsinx
12
sgnxarcsinarcsinx)
2
12
sgn(xarcsinxarcsinx)arcsinx)
2
dx12
sgnxx
arcsinx)
xx2
12
xlnxarcsinx)c
2
(4)
3
x2arccosx2arccosx
(x2arccosx)
xarccosx2xarccosx2
22
dx
3
222
xarccosxarccosxd(1x
x2dx
3
33
2x32222
xarccosx[1xarccosx(1x
darccosx]
33
2
32x3222
xarccosx[1xarccosx(1x)dx]
33
3
22x3x3222
xarccosx1xarccosxx)c
3333
12x32(xxxc
393
2
*17. (1)
求下列不定积分
cosu( 0)xu21
解:令ux
arccos(u21)dx
原式
sinx
1112du()du
2
222u2uuu(2u)c
(2)
utan
c
2sinxdxd(2cosx)
dx22cosx2cosx2cosx
x
2
du4
ln(2cosx)2
3u
1u22
(cosxdxdu) 22
1u1u
x
ln(2
cosx)ctan)ln(2cosx)c
21
sin2(x)
sinxcosx (3)
sinxcosxx)
4
dx
sin(x)d(x)
44sin(x)4
x)csc(x)d(x)444
1(sinxcosx)csc(x)cot(x)c244注:①sinxcosxsinxsin(x
)
2
x2sin
2
xcos
x
2
x
2sin(x)cosx)
444
②sinxcosxcos(
2
x)
cosx
2cosxx2
cosxx
2cos
2
③cosxsinxsin(x)
sinx
2
x2cos
④sinxcosx
cos(x)x) 444
2
xsin
x
2
x
2cos(x)sinx)
444
11
2sinxcosx(12sinxcosx1) 22
1
[(sinxcosx)21]由①,②代入 2
1122
[2sin(x)1]sin(x)2442 1[2cos2(x)1]cos2(x)12442
(4
)
tdtt21t21
44 4
1tt1t111111d(t)d(t)22
t22t22(t)22(t)22
tttt
1
t
1c
2c
c (5
)
4c
3xx
(6
)
ln(eln(exxxx
ln(exarcsin(ex)c
(7
)
dx2
x
2(arctan1
(arctan2]
2
1
2dx(arctan2
2xlnx(arctan2c
注:利用根代换找出
的一个原函数为F(x)
x
ttt2111
22tdt222(12)dtt1t1t1
2(tarctant)carctanc
x01
18.设f(x)x10x1,求f(x)dx
2xx1
x0xc
1
2
解:f(x)dxxxc0x1
22
x1xc
xf(x)(1x)f(x)
x2ex
xf(x)(1x)f(x)xf(x)f(x)xf(x)
解:x2exx2ex
19.设当x0时,f(x)连续,求
f(x)xf(x)x
]edxdxxxf(x)f(x)x
exddx
xxf(x)f(x)xf(x)x
exdedx
xxx[
f(x)f(x)xf(x)x
dxdxxxx
f(x)exc
xex
22
20.设f(cosx2)sinxtanx,求f(x)
解:令ucosx2cosxu2
u24u3
sinx1(u2)u4u3 tanx2
u4u4
2
2
2
2
u24u31
f(u)u4u32u24u4
u4u4(u2)2
1
f(x)x24x4
(x2)2
2
f(x)f(x)dx[x24x4
1
]dx2
(x2)
11x32x24xc1
3x211(x36x212x88)c1
3x211(x2)3c
3x2