27 (2)

Biblioteczka Opracowań Matematycznych

174/

Jx ² ln|x|rfx =

175/

u = ln|x| du ——    .    dr i    i

_A | , _toM+f ,    ,„_{W +}c

dv = — v = —

u = ln|x|

dv = (4 + 3xf v= J(l6 + 24x + 9x²)ix = 16x + 12x² +3x^J

J(4 + 3xf ln|.tkx =

= In|x|(l6x+I2x² + 3x³)- J(l6+12x+3x²)& = (3x⁵+12x² + 16x)ln|x|-l6x-6x² -x³ +C

176/

Jx³ ln(.x^; + 3}& =

u = ln(x² + 3) du = -y dv = x^,dx

2 xdx

x'+3

x*

= T'^^-ł^=T^lnl-⁺³|-

.iP-3_I+*_'W.£l|₁^_ł3)-i-₊^-?f^.^ln(^_ł3)-^-5Mx'₊3)₊C

2\    x²+3J    *    8    4    2V+3 4 ^    ' 8    4    4    '

=x—f₃^r*=^——L—=—fjx—L)+c In 3 ln 3    ln 3 ln 3 ln 3 ln 3 v ln 3 J

= rt+?)A ₌ f-s//²+3x// = —lnl +Vf^J + 3| + -fVr² + 3+r = ^J /V^T3 ^J    21    12

177/

Jx3‘atc =

178/

u ⁼ x du = dx

y

dv = V dx v = -

ln 3

C x²dx

^JV^T

x²-3 = r

X=V/² + 3

, tdt tdt

dx = — = -

*    V/² +3

= -^ ln|x + -Jx²-3 + xVx² -3 + C

2 I 179/

dx

f dx f ax Jx⁴+3x² " ^Jx²(x²+3)^:

1 A B Cx+D x³(A + C)+x²(B + D)+ 3.4x + 3B

: = /

x X* x +3 A=0

*4

c=o

I

,4 + C = 0 5 + Z> = 0 3/4 = 0 35 = 1    D = -j

Wracając do całki:

xV+lj~

1 rdx 1 r dx    1 V3    X _

/ = - —— —-=----arclg —j= + C

3^JX* 3 ^J X +3

180/

r dx r    dx __ V2 r dx _ J2_ r dx    1 r c

² x' - *³ - 2 J (_x² +1 \x --Jl \x + yfl)~ 2 'x-Jl 2 'x + V2 2^x² Powyższe współczynniki otrzymano z rozkładu podcałkowej funkcji wymiernej. A zatem:

+ 1

J ₄    ^ ln|jc - a/2| ln|.v + V2|-arc/gr + C

dx

¹ x* -x* -2

181/

r ^

^J(2jr + lXl + V2.x+lJ

V2.

2.r +1 = /' 2<£r = 2rt/r

bfcrhfa-

= ln|V2x + l|-ln|l + Vlr+l| + C

182/

cos .v<ir

r cos x , r cos xax    r

-^dx= -1—;-5—t= -

^Jcos3.v ^Jcosx^cos‘x-3sin‘.vj ^Jl-sii

dx

r:-h

dx

sin~x-3sin'x ^Jl-4sin'x

tgx = i

1+r²

sin x = I

! + /² A

dt

_ f 1 + /² - f ^dt _ ^    4 r    ^JI-3/²

= /

1 + /²

1 — 3/²    |-V3/ ] + V3/

fi i{aS-b43)+A+B

1-3/

{

/=- [ ^dtr +- f ^d,r =--lnll-V3/| + —T=ln|l + V3/| + C =

2^J1-V3/ 2^J1+V3/    2V3 ^1    1 2V3 ^1    1

1 + \^l3tgx

Aj3-Bj3=0 .4 + fi = I

2v/3

ln

l-V3/gx

+ C

B = — 2

-53-