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Math 191 Lecture Notes Date
§7.4 Partial Fractions
2
x+ 1+
3
x� 2
Ex1: Use Partial Fractions to find the following integralZ
5x� 1
x2 � x� 2dx
1
LCD = ( Xt 1) ( X - 2)
×÷I÷+÷I÷2X-4t3×+3_ =5×-1×2
- × - 2 XZ . X - 2
= ) ×÷d× + fxtzdx
= 2) d¥,
+ 3) ¥2w=x -2
dw=d×He*HA
du=d×= 2 f dud + 3 fdw-w
= z lnlul t 3 lnlwl t C
= 2 In 1×+11+3In 1×-21 + C✓
= In /(xH)2(x. 2)31 t c
Ex2: Use Partial Fractions to find the following integral
Z 1
0
x� 4
x2 � 5x+ 6dx
2
•
8 X -4
¥-52 ,
/ ⇒x.FI?iIxI*IIf ×=2 ⇒ - B = - 2 ⇒ B@it X= 3 ⇒ A = - 1 ⇒ A=@
. I + S÷zd×
= -In 1×-31 t
In1×-21
= ml 't¥l+c
Ex3: Use Partial Fractions to find the following integral
Zx2 � x� 6
x2 + 3xdx
3
3. x. +3 ) as + PI =HE
X 1×2+3 )
⇒ A ( × 't 3 ) + ( Bxtc ) × = ×'
. x - 6
⇐. ⇒
axn.it?EEtEEIII.E€
ftzdx + |3×xI÷d× u=x2t3
⇒ -52¥ + 3f÷×+dy.fd×÷ ,
du=2×d×
= .zinixl + 3. tzmhi " ' '
tbta"
#g.gg#taanxa
)
Ex4: Use Partial Fractions to find the following integral
Zx3 + 6x� 2
x4 + 6x2dx
4
AXTB
0×4×2+6) f
¥
÷ + BE +' I÷e=xIII÷ ,
AXCX 2+6 ) + B ( ×2t6 ) + (C x + D) X2= X 3+6×-2
€o⇒ 613=-2 ⇒ B=@)
A X3
+ 6 Ax + 13×2+613 + C X3
+ DXZ
((1/3) ) At c= 1
( cxz ) )Bt D= 0 ⇒ D=@
(( X ) )6 A = 6 ⇒ A=@ ⇒ @
s¥s¥ts*lnlxltf.txtfozftai.lt#
Ex5: Use Partial Fractions to find the following integral
Zx5 + x� 1
x3 + 1
5
1/2×3+1115+0×4+0×3+0×2 + × - l
f. X5 + + ×2
- ( X2 . xti )=
- 0 - xztx - 1
fxzdx + f -×2¥ dx L×3tl
- ( ×2 . xtl )(
xtD(x2#)
fxzdx - fdxxtl
×z3-ln1xH@
Ex6: Use Partial Fractions to find the following integral
Zdx
x1/3 + 1
6
U = x'S
⇒ x=U3
f 2¥ ⇒ dx= 3U2 duutl
- 1
3) n÷,
du u+iu%+ou+u2+ n
=
0 - U + 0
}[ Sunda + 5¥ ] If
3 [ u÷ .
u + lulutil ] + c
0
zedo.r3xlbi.lu/x43t
Ex7: Use Partial Fractions to find the following integral
Zdx
ex + 1dx
7
a. u=e×
du=e×d×I
⇒ dx=d1
fututnex
=d÷
fdu_UIUH )
g÷ + BE,=u÷u+ . )
ACUH ) + Bln ) =1
fdtu - ) # u=o⇒ @,u=t⇒B=@
:| ul -lnlutll + C
lul # Itc
⇒mle¥d@
§7.5 Strategy for integration
Ex1: Find the following integralZ
x sin x cos xdx
Ex2: Find the following integralZ
xp
x(x2p + 1)dx
8
Recall : sin 20 = Zsino cost
⇒ sinocoso = sing2
= fx sing DXu=× ,
dv=sin2×
= lzfxsinzx DX du=dx ,V= - tzco ) 2×
IBIb
=tz[ xftzcoszx ) - f-tzcossxdx]= - ÷
,XCOSZX t 1
.
Sind+ (
4 2
= - ¥Xcos2x + gt sin 2x + C
ZP
@ ← du
u=×p → u2=×
du=pxP' '
DX
t.ua.÷ =mIp X
¥ tar 'C a) tc
tptai '( xp ) t C
Ex3: Find the following integral
Zln x
xp
1 + (ln x)2dx
9
U : lnx ⇒ du=d×X
= SUE dw
now let w=1 + we
⇒ dW= zudw
= tz ) Feww
'
'2=w3 ". }
= 1 . } W"2
+ C
Z
= tg ( ituz )"2
+ c
= ! ( e + ( Inx ) 2)" 2
+ C
NOTE : You can also use trig substitution .
Ex4: Find the following integral
Ze1/x
x2dx
Ex5: Find the following integralZ
sin 6x cos 3xdx
10
- I
u= XL
Let u= tx ⇒ du= -tzzdx=
- fe"
du
U
= . e + C
Yx= - e tc
Ex6: Find the following integralZ p
1 + exdx
11
Let u= Tex
⇒ U2= 1 + ex →e×=u±1
⇒ zudu .
.
e×d×
⇒ dx= 2UdU_ /ex/
⇒ dx=2uY÷
!fu(2u÷Ddu= 2fuu÷idu weutfouto
µ2 - 1
÷ �1�
=z[jiau+fdu÷]-F.
-
- E. + In=.ua#td*eEnyIIIIEL=2ut1n1n.i1
- In Intl /
=m+hluu¥tiF×+nlF¥÷ltc
NOTE : you can also solve the
problem using trigsubstitution .
Ex7: Find the following integral
Zx+ sin�1 xp
1� x2dx
12
=f¥*dx+fsiY¥÷A B
A ⇒ ) ,÷n*dx Let x= Sino ⇒ dx=cosod0
I
⇒ g sin0t0d0# ⇒ a= fsinodl ×
the ' 0
# = - co > Otc ofxz
= -
Fx2e-
+ c
. .. -
- - . --
-.
..
B= fsifx.dxtetuI.am?d#x.
⇒ fudu⇒ u2
ztc
= ( sin''x)2=
+ c
