高等數學工-高數答案

A1lnxdx

xlnxxC2xe2xdx

1xe2x1e2xC 3xf(x)dx

xf(xf(xC4f(x滿足xf(x)dxarctanxC，則f(x)dx

lnx12ln(x21 5、若sinxf(x的原函數，則xf(x)dx=cosx2sinxxCx6u(xv(x)dx11

u(x)v(x)v(x)u(x)dx x7 x

解： dxx2t2xx

1

d(t22)=

1

dt=2

t

x=2t2lnt1C=xxlnxx

2

18、 lnx

lnx

lnx

ln

dx=(lnx1)dx=

xd(lnx1)=

x2dx=x9、xsinxcos 解：xsinxcosxdx=1xsin2xdx=1xd(cos2x)=1xcos2x1cos =1xcos2x1sin2x 10excos3x解：excos3xdxcos3xdexexcos3xexdcos3xexcos3x3=excos3x3sin3xdex=excos3x3exsin3x3exdsin3x=excos3x3exsin3x9excos3xdxexcos3xdxex(cos3x3sin3x10B3x1、 3x

解： 3x3x

x

t2

3dt6=3

5tt2t

dt=

1tx1=6(t2t11)dt=2t33t26t6lnt1C=x1t

33x66x6

6162、x2ex 解：x2exdx=x2dex=x2exexdx2=x2ex2xexdx=x2ex2=x2ex2xex2exdx=x2ex2xex2ex3、

x

解

xdxx

costdt2=2tcostdt=2tdsint=2tsint2sin==2tsint2costC=2x x2 x4

解 dx=1(1 )dx=1lnx1ln(x22)x(x2

x(x2 2 x2 5、若f(x)dxarctan2xC，求xf(x)dx。解：f(x)(arctan2x) ，xf(x)dx=

x dx=1 d(14x2)=1ln(14x2)Cex6、 ex

1ex解： ex

ex1t2

11dln(t2tttex1ex11

41 dtt

(t1

1)dt=t

C= 7、ln(x21

解ln(x21dxxln(x21xdln(x21)xln(x21

x21=xln(x21)

x2

)dx=xln(x21)2x2arctanx 8、 x1 x 解： x dxx4x

x (x2)

x21dx (x2)

x (x2)

(x2)9 d[(x2)2

x

x=2

(x2)2

x3

x2

ln[(x2)29] 1 3

Cx21、 x2xx2x 2tanx2x 2tanx12112

d(2tant)

sec212tantsec12

dt

sint12tant2sec12 12tant2sec12

lncsctcottC1212x22 xx2

F(xf(xx0時，F(x)f(x)

2(1

F(0)1，F(x)0f(x

F

f(x，所以F(x)F(x2(1x)2F(x)F(x)dx2(1x)2F(x)dF(x)1xexd1=

11d(xex)=

1exdx

1exC 1

21

21

11

21 1 1

即