Šis straipsnis yra apie racionaliųjų funkcijų integravimą .
keisti
∫
d
x
a
x
2
+
b
x
+
c
=
∫
4
a
d
x
4
a
2
x
2
+
4
a
b
x
+
b
2
+
4
a
c
−
b
2
=
4
a
∫
d
x
(
2
a
x
+
b
)
2
+
(
4
a
c
−
b
2
)
=
{\displaystyle \int {dx \over ax^{2}+bx+c}=\int {4a\;dx \over 4a^{2}x^{2}+4abx+b^{2}+4ac-b^{2}}=4a\int {dx \over (2ax+b)^{2}+(4ac-b^{2})}=}
=
4
a
2
a
∫
d
u
u
2
+
s
=
2
s
arctan
u
s
+
C
=
2
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
+
C
,
{\displaystyle ={4a \over 2a}\int {du \over u^{2}+s}={2 \over {\sqrt {s}}}\arctan {u \over {\sqrt {s}}}+C={2 \over {\sqrt {4ac-b^{2}}}}\arctan {2ax+b \over {\sqrt {4ac-b^{2}}}}+C,}
u
=
2
a
x
+
b
;
{\displaystyle u=2ax+b;}
d
u
=
2
a
d
x
;
{\displaystyle du=2adx;}
4
a
c
−
b
2
=
s
;
{\displaystyle 4ac-b^{2}=s;}
∫
M
x
+
N
a
x
2
+
b
x
+
c
d
x
=
4
a
∫
M
x
+
N
(
2
a
x
+
b
)
2
+
(
4
a
c
−
b
2
)
d
x
=
4
a
2
a
∫
M
u
−
b
2
a
+
N
u
2
+
s
d
u
=
{\displaystyle \int {Mx+N \over ax^{2}+bx+c}dx=4a\int {Mx+N \over (2ax+b)^{2}+(4ac-b^{2})}dx={4a \over 2a}\int {M{\frac {u-b}{2a}}+N \over u^{2}+s}du=}
=
1
a
∫
M
u
−
M
b
+
2
a
N
u
2
+
s
d
u
=
M
a
∫
u
d
u
u
2
+
s
+
2
a
N
−
M
b
a
∫
d
u
u
2
+
s
=
{\displaystyle ={1 \over a}\int {Mu-Mb+2aN \over u^{2}+s}du={M \over a}\int {u\;du \over u^{2}+s}+{2aN-Mb \over a}\int {du \over u^{2}+s}=}
=
M
2
a
∫
d
(
u
2
+
s
)
u
2
+
s
+
2
a
N
−
M
b
a
∫
d
u
u
2
+
s
=
M
2
a
ln
|
u
2
+
s
|
+
2
a
N
−
M
b
a
s
arctan
u
s
+
C
=
{\displaystyle ={M \over 2a}\int {d(u^{2}+s) \over u^{2}+s}+{2aN-Mb \over a}\int {du \over u^{2}+s}={M \over 2a}\ln |u^{2}+s|+{2aN-Mb \over a{\sqrt {s}}}\arctan {u \over {\sqrt {s}}}+C=}
=
M
2
a
ln
|
(
2
a
x
+
b
)
2
+
4
a
c
−
b
2
|
+
2
a
N
−
M
b
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
+
C
=
{\displaystyle ={M \over 2a}\ln |(2ax+b)^{2}+4ac-b^{2}|+{2aN-Mb \over a{\sqrt {4ac-b^{2}}}}\arctan {2ax+b \over {\sqrt {4ac-b^{2}}}}+C=}
=
M
2
a
ln
|
4
a
(
a
x
2
+
b
x
+
c
)
|
+
2
a
N
−
M
b
a
4
a
c
−
b
2
arctan
2
a
x
+
b
4
a
c
−
b
2
+
C
,
{\displaystyle ={M \over 2a}\ln |4a(ax^{2}+bx+c)|+{2aN-Mb \over a{\sqrt {4ac-b^{2}}}}\arctan {2ax+b \over {\sqrt {4ac-b^{2}}}}+C,}
u
=
2
a
x
+
b
;
{\displaystyle u=2ax+b;}
x
=
u
−
b
2
a
;
{\displaystyle x={u-b \over 2a};}
d
u
=
2
a
d
x
;
{\displaystyle du=2adx;}
4
a
c
−
b
2
=
s
;
{\displaystyle 4ac-b^{2}=s;}
s
>
0
;
{\displaystyle s>0;}
d
(
u
2
+
s
)
=
2
u
d
u
.
{\displaystyle d(u^{2}+s)=2udu.}
∫
A
x
+
B
a
x
2
+
b
x
+
c
d
x
=
∫
A
2
a
(
2
a
x
+
b
)
+
(
B
−
A
b
2
a
)
a
x
2
+
b
x
+
c
d
x
=
{\displaystyle \int {Ax+B \over ax^{2}+bx+c}dx=\int {{A \over 2a}(2ax+b)+(B-{Ab \over 2a}) \over ax^{2}+bx+c}dx=}
=
A
2
a
∫
2
a
x
+
b
a
x
2
+
b
x
+
c
d
x
+
(
B
−
A
b
2
a
)
∫
d
x
a
x
2
+
b
x
+
c
.
{\displaystyle ={A \over 2a}\int {2ax+b \over ax^{2}+bx+c}dx+(B-{Ab \over 2a})\int {dx \over ax^{2}+bx+c}.}
∫
d
x
a
x
2
+
b
x
+
c
=
1
a
∫
d
x
(
x
+
b
2
a
)
2
+
(
c
a
−
b
2
4
a
2
)
.
{\displaystyle \int {dx \over ax^{2}+bx+c}={1 \over a}\int {dx \over (x+{b \over 2a})^{2}+({c \over a}-{b^{2} \over 4a^{2}})}.}
a
x
2
+
b
x
+
c
=
a
(
x
2
+
b
a
x
+
c
a
)
=
a
[
x
2
+
2
b
2
a
x
+
(
b
2
a
)
2
+
c
a
−
(
b
2
a
)
2
]
=
a
[
(
x
+
b
2
a
)
2
+
(
c
a
−
b
2
4
a
2
)
]
.
{\displaystyle ax^{2}+bx+c=a(x^{2}+{b \over a}x+{c \over a})=a[x^{2}+2{b \over 2a}x+({b \over 2a})^{2}+{c \over a}-({b \over 2a})^{2}]=a[(x+{b \over 2a})^{2}+({c \over a}-{b^{2} \over 4a^{2}})].}
∫
A
x
+
B
a
x
2
+
b
x
+
c
d
x
=
∫
A
2
a
(
2
a
x
+
b
)
+
(
B
−
A
b
2
a
)
a
x
2
+
b
c
+
c
d
x
=
{\displaystyle \int {Ax+B \over {\sqrt {ax^{2}+bx+c}}}dx=\int {{A \over 2a}(2ax+b)+(B-{Ab \over 2a}) \over {\sqrt {ax^{2}+bc+c}}}dx=}
=
A
2
a
∫
2
a
x
+
b
a
x
2
+
b
x
+
c
d
x
+
(
B
−
A
b
2
a
)
∫
d
x
a
x
2
+
b
x
+
c
.
{\displaystyle ={A \over 2a}\int {2ax+b \over {\sqrt {ax^{2}+bx+c}}}dx+(B-{Ab \over 2a})\int {dx \over {\sqrt {ax^{2}+bx+c}}}.}
a
x
2
+
b
x
+
c
=
t
,
{\displaystyle ax^{2}+bx+c=t,}
(
2
a
x
+
b
)
d
x
=
d
t
,
{\displaystyle (2ax+b)dx=dt,}
∫
(
2
a
x
+
b
)
d
x
a
x
2
+
b
x
+
c
=
∫
d
t
t
=
2
t
+
C
=
2
a
x
2
+
b
x
+
c
+
C
.
{\displaystyle \int {(2ax+b)dx \over {\sqrt {ax^{2}+bx+c}}}=\int {dt \over {\sqrt {t}}}=2{\sqrt {t}}+C=2{\sqrt {ax^{2}+bx+c}}+C.}
∫
6
x
+
5
x
2
+
4
x
+
9
d
x
=
∫
6
x
+
5
(
x
+
2
)
2
+
5
d
x
=
∫
6
(
t
−
2
)
+
5
t
2
+
5
d
t
=
∫
6
t
−
7
t
2
+
5
d
t
=
3
∫
d
(
t
2
+
5
)
t
2
+
5
−
{\displaystyle \int {6x+5 \over x^{2}+4x+9}dx=\int {\frac {6x+5}{(x+2)^{2}+5}}dx=\int {6(t-2)+5 \over t^{2}+5}dt=\int {6t-7 \over t^{2}+5}dt=3\int {d(t^{2}+5) \over t^{2}+5}-}
−
7
∫
d
t
t
2
+
5
=
3
ln
(
t
2
+
5
)
−
7
5
arctan
t
5
+
C
=
3
ln
(
x
2
+
4
x
+
9
)
−
7
5
arctan
x
+
2
5
+
C
,
{\displaystyle -7\int {dt \over t^{2}+5}=3\ln(t^{2}+5)-{\frac {7}{\sqrt {5}}}\arctan {t \over {\sqrt {5}}}+C=3\ln(x^{2}+4x+9)-{7 \over {\sqrt {5}}}\arctan {x+2 \over {\sqrt {5}}}+C,}
kur
x
+
2
=
t
;
{\displaystyle x+2=t;}
x
=
t
−
2
;
{\displaystyle x=t-2;}
d
x
=
d
t
.
{\displaystyle dx=dt.}
∫
7
x
−
2
3
x
2
−
5
x
+
4
d
x
=
∫
84
x
−
24
36
x
2
−
60
x
+
48
d
x
=
∫
84
x
−
24
(
6
x
−
5
)
2
+
23
d
x
=
1
6
∫
14
u
+
70
−
24
u
2
+
23
d
u
=
{\displaystyle \int {7x-2 \over 3x^{2}-5x+4}dx=\int {84x-24 \over 36x^{2}-60x+48}dx=\int {84x-24 \over (6x-5)^{2}+23}dx={\frac {1}{6}}\int {14u+70-24 \over u^{2}+23}du=}
=
7
3
∫
u
d
u
u
2
+
23
+
23
3
∫
d
u
u
2
+
23
=
7
6
ln
|
u
2
+
23
|
+
23
3
23
arctan
u
23
d
u
+
C
=
{\displaystyle ={7 \over 3}\int {u\;du \over u^{2}+23}+{23 \over 3}\int {du \over u^{2}+23}={\frac {7}{6}}\ln |u^{2}+23|+{\frac {23}{3{\sqrt {23}}}}\arctan {u \over {\sqrt {23}}}du+C=}
=
7
6
ln
|
36
x
2
−
60
x
+
48
|
+
23
3
arctan
6
x
−
5
23
+
C
=
7
6
ln
|
3
x
2
−
5
x
+
4
|
+
23
3
arctan
6
x
−
5
23
+
C
1
,
{\displaystyle ={7 \over 6}\ln |36x^{2}-60x+48|+{{\sqrt {23}} \over 3}\arctan {6x-5 \over {\sqrt {23}}}+C={7 \over 6}\ln |3x^{2}-5x+4|+{{\sqrt {23}} \over 3}\arctan {6x-5 \over {\sqrt {23}}}+C_{1},}
kur
u
=
6
x
−
5
;
{\displaystyle u=6x-5;}
d
u
=
6
d
x
;
{\displaystyle du=6dx;}
x
=
u
+
5
6
.
{\displaystyle x={u+5 \over 6}.}
∫
5
x
−
3
x
2
+
6
x
−
40
d
x
=
∫
5
x
−
3
(
x
+
3
)
2
−
49
d
x
=
∫
5
u
−
15
−
3
u
2
−
49
d
u
=
5
∫
u
d
u
u
2
−
49
−
{\displaystyle \int {5x-3 \over x^{2}+6x-40}dx=\int {5x-3 \over (x+3)^{2}-49}dx=\int {5u-15-3 \over u^{2}-49}du=5\int {u\;du \over u^{2}-49}-}
−
18
∫
d
u
u
2
−
49
d
x
=
5
2
ln
|
u
2
−
49
|
−
18
14
ln
|
u
−
7
u
+
7
|
+
C
=
5
2
ln
|
x
2
+
6
x
−
40
|
−
9
7
ln
|
x
−
4
x
+
10
|
+
C
,
{\displaystyle -18\int {du \over u^{2}-49}dx={5 \over 2}\ln |u^{2}-49|-{18 \over 14}\ln |{u-7 \over u+7}|+C={5 \over 2}\ln |x^{2}+6x-40|-{9 \over 7}\ln |{x-4 \over x+10}|+C,}
kur
u
=
x
+
3
;
{\displaystyle u=x+3;}
d
u
=
d
x
;
{\displaystyle du=dx;}
x
=
u
−
3.
{\displaystyle x=u-3.}
∫
5
x
−
3
x
2
+
6
x
−
40
d
x
=
∫
4
(
5
x
−
3
)
d
x
4
x
2
+
24
x
+
36
−
160
−
36
=
∫
4
(
5
x
−
3
)
d
x
(
2
x
+
6
)
2
−
196
=
4
2
∫
5
u
−
6
2
−
3
u
2
−
196
d
u
=
{\displaystyle \int {5x-3 \over x^{2}+6x-40}dx=\int {4(5x-3)dx \over 4x^{2}+24x+36-160-36}=\int {4(5x-3)dx \over (2x+6)^{2}-196}={4 \over 2}\int {5{u-6 \over 2}-3 \over u^{2}-196}du=}
=
2
2
∫
5
u
−
30
−
6
u
2
−
196
d
u
=
5
∫
u
d
u
u
2
−
196
−
36
∫
d
u
u
2
−
196
=
5
2
∫
d
(
u
2
−
196
)
u
2
−
196
−
36
28
ln
|
u
−
14
u
+
14
|
+
C
=
{\displaystyle ={2 \over 2}\int {5u-30-6 \over u^{2}-196}du=5\int {u\;du \over u^{2}-196}-36\int {du \over u^{2}-196}={5 \over 2}\int {d(u^{2}-196) \over u^{2}-196}-{36 \over 28}\ln |{u-14 \over u+14}|+C=}
=
5
2
ln
|
u
2
−
196
|
−
9
7
ln
|
2
x
−
8
2
x
+
20
|
+
C
1
=
5
2
ln
|
4
(
x
2
+
6
x
−
40
)
|
−
9
7
ln
|
x
−
4
x
+
10
|
+
C
1
,
{\displaystyle ={5 \over 2}\ln |u^{2}-196|-{9 \over 7}\ln |{2x-8 \over 2x+20}|+C_{1}={5 \over 2}\ln |4(x^{2}+6x-40)|-{9 \over 7}\ln |{x-4 \over x+10}|+C_{1},}
2
x
+
6
=
u
;
{\displaystyle 2x+6=u;}
x
=
u
−
6
2
;
{\displaystyle x={u-6 \over 2};}
d
u
=
2
d
x
;
{\displaystyle du=2dx;}
d
(
u
2
−
196
)
=
2
u
d
u
.
{\displaystyle d(u^{2}-196)=2udu.}
∫
3
x
+
4
7
−
x
2
+
6
x
d
x
=
∫
3
x
+
4
16
−
(
x
−
3
)
2
d
x
=
∫
3
u
+
9
+
4
16
−
u
2
d
u
=
∫
3
u
d
u
16
−
u
2
+
∫
13
d
u
16
−
u
2
=
{\displaystyle \int {3x+4 \over {\sqrt {7-x^{2}+6x}}}dx=\int {3x+4 \over {\sqrt {16-(x-3)^{2}}}}dx=\int {3u+9+4 \over {\sqrt {16-u^{2}}}}du=\int {3u\;du \over {\sqrt {16-u^{2}}}}+\int {13du \over {\sqrt {16-u^{2}}}}=}
=
−
3
16
−
u
2
+
13
arcsin
u
4
+
C
=
−
3
7
−
x
2
+
6
x
+
13
arcsin
x
−
3
4
+
C
,
{\displaystyle =-3{\sqrt {16-u^{2}}}+13\arcsin {u \over 4}+C=-3{\sqrt {7-x^{2}+6x}}+13\arcsin {x-3 \over 4}+C,}
u
=
x
−
3
;
{\displaystyle u=x-3;}
d
u
=
d
x
;
{\displaystyle du=dx;}
x
=
u
+
3
;
{\displaystyle x=u+3;}
d
(
16
−
u
2
)
=
−
2
u
d
u
.
{\displaystyle d(16-u^{2})=-2u\,du.}
∫
(
x
+
3
)
d
x
x
2
−
2
x
−
5
=
∫
1
2
(
2
x
−
2
)
+
(
3
+
1
2
2
)
x
2
−
2
x
−
5
d
x
=
1
2
∫
(
2
x
−
2
)
d
x
x
2
−
2
x
−
5
+
4
∫
d
x
x
2
−
2
x
−
5
=
{\displaystyle \int {(x+3)dx \over x^{2}-2x-5}=\int {{1 \over 2}(2x-2)+(3+{1 \over 2}2) \over x^{2}-2x-5}dx={1 \over 2}\int {(2x-2)dx \over x^{2}-2x-5}+4\int {dx \over x^{2}-2x-5}=}
=
1
2
∫
d
(
x
2
−
2
x
−
5
)
x
2
−
2
x
−
5
+
4
∫
d
x
(
x
−
1
)
2
−
6
=
1
2
ln
|
x
2
−
2
x
−
5
|
+
4
1
6
ln
|
6
−
(
x
−
1
)
6
+
(
x
−
1
)
|
+
C
.
{\displaystyle ={1 \over 2}\int {d(x^{2}-2x-5) \over x^{2}-2x-5}+4\int {dx \over (x-1)^{2}-6}={1 \over 2}\ln |x^{2}-2x-5|+4{1 \over {\sqrt {6}}}\ln |{{\sqrt {6}}-(x-1) \over {\sqrt {6}}+(x-1)}|+C.}
∫
d
x
2
x
2
+
8
x
+
20
=
1
2
∫
d
x
x
2
+
4
x
+
10
=
1
2
∫
d
x
1
[
(
x
+
4
2
⋅
1
)
2
+
(
10
1
−
4
2
4
⋅
1
2
)
]
=
{\displaystyle \int {dx \over 2x^{2}+8x+20}={1 \over 2}\int {dx \over x^{2}+4x+10}={1 \over 2}\int {dx \over 1[(x+{4 \over 2\cdot 1})^{2}+({10 \over 1}-{4^{2} \over 4\cdot 1^{2}})]}=}
=
1
2
∫
d
x
(
x
2
+
4
x
+
4
)
+
(
10
−
4
)
=
1
2
∫
d
x
(
x
+
2
)
2
+
6
=
1
2
∫
d
t
t
2
+
6
=
1
2
1
6
arctan
t
6
+
C
=
{\displaystyle ={1 \over 2}\int {dx \over (x^{2}+4x+4)+(10-4)}={1 \over 2}\int {dx \over (x+2)^{2}+6}={1 \over 2}\int {dt \over t^{2}+6}={1 \over 2}{1 \over {\sqrt {6}}}\arctan {t \over {\sqrt {6}}}+C=}
=
1
2
6
arctan
x
+
2
6
+
C
,
{\displaystyle ={1 \over 2{\sqrt {6}}}\arctan {x+2 \over {\sqrt {6}}}+C,}
x
+
2
=
t
,
{\displaystyle x+2=t,}
d
x
=
d
t
.
{\displaystyle dx=dt.}
∫
5
x
+
3
x
2
+
4
x
+
10
d
x
=
∫
5
2
(
2
x
+
4
)
+
(
3
−
10
)
x
2
+
4
x
+
10
d
x
=
{\displaystyle \int {5x+3 \over {\sqrt {x^{2}+4x+10}}}dx=\int {{5 \over 2}(2x+4)+(3-10) \over {\sqrt {x^{2}+4x+10}}}dx=}
=
5
2
∫
2
x
+
4
x
2
+
4
x
+
10
d
x
−
7
∫
d
x
(
x
+
2
)
2
+
6
=
{\displaystyle ={5 \over 2}\int {2x+4 \over {\sqrt {x^{2}+4x+10}}}dx-7\int {dx \over {\sqrt {(x+2)^{2}+6}}}=}
=
5
2
∫
d
(
x
2
+
4
x
+
10
)
x
2
+
4
x
+
10
−
7
ln
|
x
+
2
+
(
x
+
2
)
2
+
6
|
+
C
=
{\displaystyle ={5 \over 2}\int {d(x^{2}+4x+10) \over {\sqrt {x^{2}+4x+10}}}-7\ln |x+2+{\sqrt {(x+2)^{2}+6}}|+C=}
=
5
x
2
+
4
x
+
10
−
7
ln
|
x
+
2
+
x
2
+
4
x
+
10
|
+
C
,
{\displaystyle =5{\sqrt {x^{2}+4x+10}}-7\ln |x+2+{\sqrt {x^{2}+4x+10}}|+C,}
d
(
x
2
+
4
x
+
10
)
=
(
2
x
+
4
)
d
x
;
{\displaystyle d(x^{2}+4x+10)=(2x+4)dx;}
d
(
x
+
2
)
=
d
x
.
{\displaystyle d(x+2)=dx.}
keisti
Teorema . Jeigu racionali funkcija
R
(
x
)
Q
(
x
)
{\displaystyle {R(x) \over Q(x)}}
Q(x) pateiktas pavidale
Q
(
x
)
=
A
(
x
−
α
)
r
(
x
−
β
)
s
.
.
.
(
x
2
+
2
p
x
+
q
)
g
(
x
2
+
2
u
x
+
v
)
h
.
.
.
,
{\displaystyle Q(x)=A(x-\alpha )^{r}(x-\beta )^{s}...(x^{2}+2px+q)^{g}(x^{2}+2ux+v)^{h}...,}
kur
α
,
β
,
.
.
.
,
p
,
q
,
u
,
v
.
.
.
{\displaystyle \alpha ,\;\beta ,\;...,p,\;q,\;u,\;v\;...}
r, s, ..., g, h , ... - naturalieji skaičiai, tai šitą funkiciją galima vieninteliu budu pateikti pavidale
R
(
x
)
Q
(
x
)
=
A
1
(
x
−
α
)
+
A
2
(
x
−
α
)
2
+
.
.
.
+
A
r
(
x
−
α
)
r
+
B
1
(
x
−
β
)
+
B
2
(
x
−
β
)
2
+
.
.
.
+
B
s
(
x
−
β
)
s
+
.
.
.
+
{\displaystyle {R(x) \over Q(x)}={A_{1} \over (x-\alpha )}+{A_{2} \over (x-\alpha )^{2}}+...+{A_{r} \over (x-\alpha )^{r}}+{B_{1} \over (x-\beta )}+{B_{2} \over (x-\beta )^{2}}+...+{B_{s} \over (x-\beta )^{s}}+...+}
+
M
1
x
+
N
1
x
2
+
2
p
x
+
q
+
M
2
x
+
N
2
(
x
2
+
2
p
x
+
q
)
2
+
.
.
.
+
M
g
x
+
N
g
(
x
2
+
2
p
x
+
q
)
g
+
K
1
x
+
L
1
x
2
+
2
u
x
+
v
+
K
2
x
+
L
2
(
x
2
+
2
u
x
+
v
)
2
+
.
.
.
+
K
h
x
+
L
h
(
x
2
+
2
u
x
+
v
)
h
+
.
.
.
{\displaystyle +{M_{1}x+N_{1} \over x^{2}+2px+q}+{M_{2}x+N_{2} \over (x^{2}+2px+q)^{2}}+...+{M_{g}x+N_{g} \over (x^{2}+2px+q)^{g}}+{K_{1}x+L_{1} \over x^{2}+2ux+v}+{K_{2}x+L_{2} \over (x^{2}+2ux+v)^{2}}+...+{K_{h}x+L_{h} \over (x^{2}+2ux+v)^{h}}+...}
kur
A
1
{\displaystyle A_{1}}
A
2
{\displaystyle A_{2}}
A
r
{\displaystyle A_{r}}
B
1
{\displaystyle B_{1}}
B
2
{\displaystyle B_{2}}
B
s
{\displaystyle B_{s}}
M
1
{\displaystyle M_{1}}
N
1
{\displaystyle N_{1}}
M
2
{\displaystyle M_{2}}
N
2
{\displaystyle N_{2}}
M
g
{\displaystyle M_{g}}
N
g
{\displaystyle N_{g}}
K
1
{\displaystyle K_{1}}
L
1
{\displaystyle L_{1}}
K
2
{\displaystyle K_{2}}
L
2
{\displaystyle L_{2}}
K
h
{\displaystyle K_{h}}
L
h
{\displaystyle L_{h}}
Daugianaris (polinomas) Q (x ) gali būti pateiktas neišskaidytas dauginamaisiais. Tada reikės surasti daugianario Q (x ) visas realiąsias ir menamas šaknis ir žinoti jų kartotinumą. Šiame
Q
(
x
)
=
A
(
x
−
α
)
r
(
x
−
β
)
s
.
.
.
(
x
2
+
2
p
x
+
q
)
g
(
x
2
+
2
u
x
+
v
)
h
.
.
.
{\displaystyle Q(x)=A(x-\alpha )^{r}(x-\beta )^{s}...(x^{2}+2px+q)^{g}(x^{2}+2ux+v)^{h}...}
daugianaryje
α
,
β
,
.
.
.
{\displaystyle \alpha ,\;\beta ,\;...}
Q (x ) šaknys atitinkamai kartotinumo r , s , ... . O reiškiniai
x
2
+
2
p
x
+
q
=
(
x
−
z
1
)
(
x
−
z
1
¯
)
,
x
2
+
2
u
x
+
v
=
(
x
−
z
2
)
(
x
−
z
2
¯
)
,
.
.
.
{\displaystyle x^{2}+2px+q=(x-z_{1})(x-{\overline {z_{1}}}),\;x^{2}+2ux+v=(x-z_{2})(x-{\overline {z_{2}}}),...\;}
Q (x ) menamas jungtines šaknis
z
1
{\displaystyle z_{1}}
z
1
¯
,
{\displaystyle {\overline {z_{1}}},\;}
z
2
{\displaystyle z_{2}}
z
2
¯
,
.
.
.
,
{\displaystyle {\overline {z_{2}}},\;...,}
g , h , ... . Pavyzdžiui, kompleksiniam skaičiui
z
=
3
+
4
i
{\displaystyle z=3+4i}
z
¯
=
3
−
4
i
.
{\displaystyle {\overline {z}}=3-4i.}
keisti
Tegu turime polinomą
Q
(
x
)
=
(
x
−
1
)
3
(
x
−
2
)
.
{\displaystyle Q(x)=(x-1)^{3}(x-2).}
Q (x ) vienos šaknies (x=1) kartotinumas yra 3, o kitos šaknies (x=2) kartotinumas yra 1 (paprasta šankis).
Q
(
x
)
=
(
x
−
1
)
3
(
x
−
2
)
=
(
x
3
−
3
x
2
+
3
x
−
1
)
(
x
−
2
)
=
x
4
−
3
x
3
+
3
x
2
−
x
−
2
x
3
+
6
x
2
−
6
x
+
2
=
x
4
−
5
x
3
+
9
x
2
−
7
x
+
2.
{\displaystyle Q(x)=(x-1)^{3}(x-2)=(x^{3}-3x^{2}+3x-1)(x-2)=x^{4}-3x^{3}+3x^{2}-x-2x^{3}+6x^{2}-6x+2=x^{4}-5x^{3}+9x^{2}-7x+2.}
Taigi,
(
x
−
1
)
3
(
x
−
2
)
=
x
4
−
5
x
3
+
9
x
2
−
7
x
+
2.
{\displaystyle (x-1)^{3}(x-2)=x^{4}-5x^{3}+9x^{2}-7x+2.}
Kad surasti polinomo
Q
(
x
)
=
x
4
−
5
x
3
+
9
x
2
−
7
x
+
2
{\displaystyle Q(x)=x^{4}-5x^{3}+9x^{2}-7x+2}
n -ta polinomo Q(x) išvestinė su x=1 reikšme nelygi 0, o n-1, n-2 ir visos mažesnės eilės nei n išvestinės lygios nuliui. Tada polinomo Q(x) šaknies x=1 kartotinumas yra n . Įrodymas per pavyzdį.Kai surasime polinomo Q(x) trečios eilės išvestinę ir įstatysime x=1, tai
Q
‴
(
1
)
{\displaystyle Q'''(1)}
Q
′
(
x
)
=
[
(
x
−
1
)
3
(
x
−
2
)
]
′
=
(
x
4
−
5
x
3
+
9
x
2
−
7
x
+
2
)
′
,
{\displaystyle Q'(x)=[(x-1)^{3}(x-2)]'=(x^{4}-5x^{3}+9x^{2}-7x+2)',}
Q
′
(
x
)
=
[
3
(
x
−
1
)
2
(
x
−
2
)
+
(
x
−
1
)
3
]
=
(
4
x
3
−
15
x
2
+
18
x
−
7
)
;
{\displaystyle Q'(x)=[3(x-1)^{2}(x-2)+(x-1)^{3}]=(4x^{3}-15x^{2}+18x-7);}
matome, kad Q'(1)=0;
Q
″
(
x
)
=
[
3
(
x
−
1
)
2
(
x
−
2
)
+
(
x
−
1
)
3
]
′
=
(
4
x
3
−
15
x
2
+
18
x
−
7
)
′
,
{\displaystyle Q''(x)=[3(x-1)^{2}(x-2)+(x-1)^{3}]'=(4x^{3}-15x^{2}+18x-7)',}
Q
″
(
x
)
=
[
6
(
x
−
1
)
(
x
−
2
)
+
3
(
x
−
1
)
2
(
x
−
2
)
+
3
(
x
−
1
)
2
]
=
(
12
x
2
−
30
x
+
18
)
;
{\displaystyle Q''(x)=[6(x-1)(x-2)+3(x-1)^{2}(x-2)+3(x-1)^{2}]=(12x^{2}-30x+18);}
ir dabar matome, kad
Q
″
(
1
)
=
0
;
{\displaystyle Q''(1)=0;}
Q
‴
(
x
)
=
[
6
(
x
−
1
)
(
x
−
2
)
+
3
(
x
−
1
)
2
(
x
−
2
)
+
3
(
x
−
1
)
2
]
′
=
(
12
x
2
−
30
x
+
18
)
′
,
{\displaystyle Q'''(x)=[6(x-1)(x-2)+3(x-1)^{2}(x-2)+3(x-1)^{2}]'=(12x^{2}-30x+18)',}
Q
‴
(
x
)
=
[
6
(
x
−
2
)
+
6
(
x
−
1
)
+
6
(
x
−
1
)
(
x
−
2
)
+
3
(
x
−
1
)
2
+
6
(
x
−
1
)
]
=
(
24
x
−
30
)
;
{\displaystyle Q'''(x)=[6(x-2)+6(x-1)+6(x-1)(x-2)+3(x-1)^{2}+6(x-1)]=(24x-30);}
na o dabar
Q
‴
(
1
)
≠
0.
{\displaystyle Q'''(1)\neq 0.}
Q
‴
(
1
)
=
[
6
(
1
−
2
)
+
6
(
1
−
1
)
+
6
(
1
−
1
)
(
x
−
2
)
+
3
(
1
−
1
)
2
+
6
(
1
−
1
)
]
=
−
6
;
{\displaystyle Q'''(1)=[6(1-2)+6(1-1)+6(1-1)(x-2)+3(1-1)^{2}+6(1-1)]=-6;}
Q
‴
(
1
)
=
(
24
⋅
1
−
30
)
=
−
6.
{\displaystyle Q'''(1)=(24\cdot 1-30)=-6.}
Vadinasi polinomo Q(x) šaknies x=1 kartotinumas yra 3. Įrodymo esmė, kad imant išvestines, narys (x-1) prie tam tikros išvestinės eilės n pranyks (ir liks tik (x-2)) ir
Q
(
n
)
(
1
)
{\displaystyle Q^{(n)}(1)}
Analogiškai galima įrodyti, kai polinomas išskaidytas daugikliais su bet kokiais laipsniais. Tereikia taikyti funkcijų sandaugos diferencijavimo taisyklę
[
u
(
x
)
⋅
v
(
x
)
]
′
=
u
′
(
x
)
v
(
x
)
+
u
(
x
)
v
′
(
x
)
.
{\displaystyle [u(x)\cdot v(x)]'=u'(x)v(x)+u(x)v'(x).}
u
(
x
)
=
(
x
−
1
)
3
,
v
(
x
)
=
x
−
2.
{\displaystyle u(x)=(x-1)^{3},\;\;v(x)=x-2.}
Kai turime 3 funkcijas nuo x , tai taikome tą pačią dviejų funkcijų diferencijavimo taisyklę (
[
u
(
x
)
⋅
v
(
x
)
]
′
=
u
′
(
x
)
v
(
x
)
+
u
(
x
)
v
′
(
x
)
{\displaystyle [u(x)\cdot v(x)]'=u'(x)v(x)+u(x)v'(x)}
x taikome, lyg u(x) ir v(x) sandauga būtų g(x) funkcija (g(x)=u(x)v(x)), o w(x) lyg būtų atskira funkcija:
Q
′
(
x
)
=
[
u
v
w
]
′
=
(
u
v
)
′
w
+
u
v
w
′
=
(
u
′
v
+
u
v
′
)
w
+
u
v
w
′
=
u
′
v
w
+
u
v
′
w
+
u
v
w
′
.
{\displaystyle Q'(x)=[uvw]'=(uv)'w+uvw'=(u'v+uv')w+uvw'=u'vw+uv'w+uvw'.}
Indukcijos metodu, gautume, kad
Q
′
(
x
)
=
[
u
1
u
2
u
3
.
.
.
u
m
]
′
=
u
1
′
u
2
u
3
.
.
.
u
m
+
u
1
u
2
′
u
3
.
.
.
u
m
+
u
1
u
2
u
3
′
.
.
.
u
m
+
.
.
.
+
u
1
u
2
u
3
.
.
.
u
m
′
.
{\displaystyle Q'(x)=[u_{1}u_{2}u_{3}...u_{m}]'=u_{1}'u_{2}u_{3}...u_{m}+u_{1}u_{2}'u_{3}...u_{m}+u_{1}u_{2}u_{3}'...u_{m}+...+u_{1}u_{2}u_{3}...u_{m}'.}
Yra ir kitų būdų polinomo šaknies kartotinumui surasti. Vienas budas yra toks, kad mūsų pavyzdyje polinomą Q(x) reikia dalinti iš (x-1) tiek kartų, kol
Q
(
x
)
=
x
4
−
5
x
3
+
9
x
2
−
7
x
+
2
{\displaystyle Q(x)=x^{4}-5x^{3}+9x^{2}-7x+2}
n -tas kartas, kai (x-1) padalins Q(x) be liekanos ir reikš šaknies x=1 kartotinumą, lygų n .
keisti
∫
x
5
+
x
4
−
8
x
3
−
4
x
d
x
=
∫
[
x
2
+
x
+
4
+
4
x
2
+
16
x
−
8
x
(
x
+
2
)
(
x
−
2
)
]
d
x
=
x
3
3
+
x
2
2
+
4
x
+
4
∫
x
2
+
4
x
−
2
x
(
x
+
2
)
(
x
−
2
)
d
x
.
{\displaystyle \int {\frac {x^{5}+x^{4}-8}{x^{3}-4x}}dx=\int [x^{2}+x+4+{\frac {4x^{2}+16x-8}{x(x+2)(x-2)}}]dx={\frac {x^{3}}{3}}+{\frac {x^{2}}{2}}+4x+4\int {\frac {x^{2}+4x-2}{x(x+2)(x-2)}}dx.}
x
2
+
4
x
−
2
x
(
x
+
2
)
(
x
−
2
)
=
A
x
+
B
x
+
2
+
C
x
−
2
.
{\displaystyle {\frac {x^{2}+4x-2}{x(x+2)(x-2)}}={\frac {A}{x}}+{\frac {B}{x+2}}+{\frac {C}{x-2}}.}
x
(
x
+
2
)
(
x
−
2
)
,
{\displaystyle x(x+2)(x-2),}
x
2
+
4
x
−
2
=
A
(
x
+
2
)
(
x
−
2
)
+
B
x
(
x
−
2
)
+
C
x
(
x
+
2
)
=
A
(
x
2
−
4
)
+
B
(
x
2
−
2
x
)
+
C
(
x
2
+
2
x
)
=
x
2
(
A
+
B
+
C
)
+
x
(
−
2
B
+
2
C
)
−
4
A
.
{\displaystyle x^{2}+4x-2=A(x+2)(x-2)+Bx(x-2)+Cx(x+2)=A(x^{2}-4)+B(x^{2}-2x)+C(x^{2}+2x)=x^{2}(A+B+C)+x(-2B+2C)-4A.}
A
+
B
+
C
=
1
,
{\displaystyle A+B+C=1,}
−
2
B
+
2
C
=
4
,
{\displaystyle -2B+2C=4,}
−
4
A
=
−
2.
{\displaystyle -4A=-2.}
Iš sistemos randame:
A
=
1
2
;
B
=
−
3
4
;
C
=
5
4
.
{\displaystyle A={\frac {1}{2}};\;B=-{\frac {3}{4}};\;C={\frac {5}{4}}.}
4
∫
x
2
+
4
x
−
2
x
(
x
+
2
)
(
x
−
2
)
d
x
=
4
(
1
2
∫
d
x
x
−
3
4
∫
d
x
x
+
2
+
5
4
∫
d
x
x
−
2
)
=
{\displaystyle 4\int {\frac {x^{2}+4x-2}{x(x+2)(x-2)}}dx=4({\frac {1}{2}}\int {\frac {dx}{x}}-{\frac {3}{4}}\int {\frac {dx}{x+2}}+{\frac {5}{4}}\int {\frac {dx}{x-2}})=}
=
4
(
1
2
ln
|
x
|
−
3
4
ln
|
x
+
2
|
+
5
4
ln
|
x
−
2
|
)
+
C
=
ln
|
x
2
(
x
−
2
)
5
(
x
+
2
)
3
|
+
C
.
{\displaystyle =4({\frac {1}{2}}\ln |x|-{\frac {3}{4}}\ln |x+2|+{\frac {5}{4}}\ln |x-2|)+C=\ln |{\frac {x^{2}(x-2)^{5}}{(x+2)^{3}}}|+C.}
∫
x
5
+
x
4
−
8
x
3
−
4
x
d
x
=
x
3
3
+
x
2
2
+
4
x
+
ln
|
x
2
(
x
−
2
)
5
(
x
+
2
)
3
|
+
C
.
{\displaystyle \int {\frac {x^{5}+x^{4}-8}{x^{3}-4x}}dx={\frac {x^{3}}{3}}+{\frac {x^{2}}{2}}+4x+\ln |{\frac {x^{2}(x-2)^{5}}{(x+2)^{3}}}|+C.}
∫
x
d
x
x
2
−
6
x
+
12
=
∫
x
d
x
(
x
−
3
)
2
+
3
.
{\displaystyle \int {\frac {xdx}{x^{2}-6x+12}}=\int {\frac {xdx}{(x-3)^{2}+3}}.}
x
−
3
=
u
,
{\displaystyle x-3=u,}
∫
(
u
+
3
)
d
u
u
2
+
3
=
∫
u
d
u
u
2
+
3
+
3
∫
d
u
u
2
+
3
=
1
2
∫
d
(
u
2
+
3
)
u
2
+
3
+
3
∫
d
u
u
2
+
(
3
)
2
=
{\displaystyle \int {\frac {(u+3)du}{u^{2}+3}}=\int {\frac {u\;du}{u^{2}+3}}+3\int {\frac {du}{u^{2}+3}}={\frac {1}{2}}\int {\frac {d(u^{2}+3)}{u^{2}+3}}+3\int {\frac {du}{u^{2}+({\sqrt {3}})^{2}}}=}
=
1
2
ln
(
u
2
+
3
)
+
3
3
arctan
u
3
+
C
=
1
2
ln
(
x
2
−
6
x
+
12
)
+
3
arctan
x
−
3
3
+
C
.
{\displaystyle ={\frac {1}{2}}\ln(u^{2}+3)+{\frac {3}{\sqrt {3}}}\arctan {\frac {u}{\sqrt {3}}}+C={\frac {1}{2}}\ln(x^{2}-6x+12)+{\sqrt {3}}\arctan {\frac {x-3}{\sqrt {3}}}+C.}
Šį rezultatą buvo galima gauti iš karto remiantis
∫
M
x
+
N
x
2
+
p
x
+
q
d
x
=
M
2
ln
|
x
2
+
p
x
+
q
|
+
2
N
−
M
p
4
q
−
p
2
arctan
2
x
+
p
4
q
−
p
2
+
C
{\displaystyle \int {\frac {Mx+N}{x^{2}+px+q}}dx={\frac {M}{2}}\ln |x^{2}+px+q|+{\frac {2N-Mp}{\sqrt {4q-p^{2}}}}\arctan {\frac {2x+p}{\sqrt {4q-p^{2}}}}+C}
M
=
1
,
{\displaystyle M=1,}
N
=
0
,
{\displaystyle N=0,}
p
=
−
6
{\displaystyle p=-6}
q
=
12.
{\displaystyle q=12.}
∫
x
d
x
x
2
−
6
x
+
12
=
1
2
ln
(
x
2
−
6
x
+
12
)
+
−
1
⋅
(
−
6
)
48
−
36
arctan
2
x
−
6
48
−
36
+
C
=
{\displaystyle \int {\frac {xdx}{x^{2}-6x+12}}={\frac {1}{2}}\ln(x^{2}-6x+12)+{\frac {-1\cdot (-6)}{\sqrt {48-36}}}\arctan {\frac {2x-6}{\sqrt {48-36}}}+C=}
=
1
2
ln
(
x
2
−
6
x
+
12
)
+
3
arctan
x
−
3
3
+
C
.
{\displaystyle ={\frac {1}{2}}\ln(x^{2}-6x+12)+{\sqrt {3}}\arctan {\frac {x-3}{\sqrt {3}}}+C.}
∫
15
x
2
−
4
x
−
81
x
3
−
13
x
+
12
d
x
=
∫
15
x
2
−
4
x
−
81
(
x
−
3
)
(
x
+
4
)
(
x
−
1
)
d
x
=
∫
[
A
x
−
3
+
B
x
+
4
+
D
x
−
1
]
d
x
.
{\displaystyle \int {\frac {15x^{2}-4x-81}{x^{3}-13x+12}}dx=\int {\frac {15x^{2}-4x-81}{(x-3)(x+4)(x-1)}}dx=\int [{\frac {A}{x-3}}+{\frac {B}{x+4}}+{\frac {D}{x-1}}]dx.}
15
x
2
−
4
x
−
81
=
A
(
x
+
4
)
(
x
−
1
)
+
B
(
x
−
3
)
(
x
−
1
)
+
D
(
x
−
3
)
(
x
+
4
)
=
A
(
x
2
+
3
x
−
4
)
+
B
(
x
2
−
4
x
+
3
)
+
D
(
x
2
+
x
−
12
)
.
{\displaystyle 15x^{2}-4x-81=A(x+4)(x-1)+B(x-3)(x-1)+D(x-3)(x+4)=A(x^{2}+3x-4)+B(x^{2}-4x+3)+D(x^{2}+x-12).}
x laipsnių, gauname tiesinių lygčių sistemą
A
+
B
+
D
=
15
,
{\displaystyle A+B+D=15,}
3
A
−
4
B
+
D
=
−
4
,
{\displaystyle 3A-4B+D=-4,}
−
4
A
+
3
B
−
12
D
=
−
81.
{\displaystyle -4A+3B-12D=-81.}
Iš sistemos randame:
A
=
3
;
{\displaystyle A=3;}
B
=
5
;
{\displaystyle B=5;}
D
=
7.
{\displaystyle D=7.}
∫
15
x
2
−
4
x
−
81
x
3
−
13
x
+
12
d
x
=
3
∫
d
x
x
−
3
+
5
∫
d
x
x
+
4
+
7
∫
d
x
x
−
1
=
{\displaystyle \int {\frac {15x^{2}-4x-81}{x^{3}-13x+12}}dx=3\int {\frac {dx}{x-3}}+5\int {\frac {dx}{x+4}}+7\int {\frac {dx}{x-1}}=}
=
3
ln
|
x
−
3
|
+
5
ln
|
x
+
4
|
+
7
ln
|
x
−
1
|
+
C
=
ln
|
(
x
−
3
)
3
(
x
+
4
)
5
(
x
−
1
)
7
|
+
C
.
{\displaystyle =3\ln |x-3|+5\ln |x+4|+7\ln |x-1|+C=\ln |(x-3)^{3}(x+4)^{5}(x-1)^{7}|+C.}
∫
x
4
−
3
x
2
−
3
x
−
2
x
3
−
x
2
−
2
x
d
x
=
∫
[
x
+
1
−
x
+
2
x
(
x
2
−
x
−
2
)
]
d
x
;
{\displaystyle \int {\frac {x^{4}-3x^{2}-3x-2}{x^{3}-x^{2}-2x}}dx=\int [x+1-{\frac {x+2}{x(x^{2}-x-2)}}]dx;}
x
+
2
x
(
x
−
2
)
(
x
+
1
)
=
A
x
+
B
x
−
2
+
D
x
+
1
;
{\displaystyle {\frac {x+2}{x(x-2)(x+1)}}={\frac {A}{x}}+{\frac {B}{x-2}}+{\frac {D}{x+1}};}
x
+
2
=
A
(
x
−
2
)
(
x
+
1
)
+
B
x
(
x
+
1
)
+
D
x
(
x
−
2
)
=
A
(
x
2
−
x
−
2
)
+
B
(
x
2
+
x
)
+
D
(
x
2
−
2
x
)
=
x
2
(
A
+
B
+
D
)
+
x
(
−
A
+
B
−
2
D
)
−
2
A
.
{\displaystyle x+2=A(x-2)(x+1)+Bx(x+1)+Dx(x-2)=A(x^{2}-x-2)+B(x^{2}+x)+D(x^{2}-2x)=x^{2}(A+B+D)+x(-A+B-2D)-2A.}
A
+
B
+
D
=
0
;
{\displaystyle A+B+D=0;}
−
A
+
B
−
2
D
=
1
;
{\displaystyle -A+B-2D=1;}
−
2
A
=
2
;
{\displaystyle -2A=2;}
A
=
−
1
;
{\displaystyle A=-1;}
B
=
2
3
;
{\displaystyle B={\frac {2}{3}};}
D
=
1
3
.
{\displaystyle D={\frac {1}{3}}.}
∫
x
4
−
3
x
2
−
3
x
−
2
x
3
−
x
2
−
2
x
d
x
=
{\displaystyle \int {\frac {x^{4}-3x^{2}-3x-2}{x^{3}-x^{2}-2x}}dx=}
=
∫
(
x
+
1
)
d
x
+
∫
d
x
x
−
2
3
∫
d
x
x
−
2
−
1
3
∫
d
x
x
+
1
=
x
2
2
+
x
+
ln
|
x
|
−
2
3
ln
|
x
−
2
|
−
1
3
ln
|
x
+
1
|
+
C
.
{\displaystyle =\int (x+1)dx+\int {\frac {dx}{x}}-{\frac {2}{3}}\int {\frac {dx}{x-2}}-{\frac {1}{3}}\int {\frac {dx}{x+1}}={\frac {x^{2}}{2}}+x+\ln |x|-{\frac {2}{3}}\ln |x-2|-{\frac {1}{3}}\ln |x+1|+C.}
∫
2
x
2
−
3
x
+
3
x
3
−
2
x
2
+
x
d
x
=
∫
[
A
x
+
B
(
x
−
1
)
2
+
D
x
−
1
]
d
x
.
{\displaystyle \int {\frac {2x^{2}-3x+3}{x^{3}-2x^{2}+x}}dx=\int [{\frac {A}{x}}+{\frac {B}{(x-1)^{2}}}+{\frac {D}{x-1}}]dx.}
2
x
2
−
3
x
+
3
=
A
(
x
−
1
)
2
+
B
x
+
D
x
(
x
−
1
)
=
A
(
x
2
−
2
x
+
1
)
+
B
x
+
D
(
x
2
−
x
)
=
x
2
(
A
+
D
)
+
x
(
−
2
A
+
B
−
D
)
+
A
.
{\displaystyle 2x^{2}-3x+3=A(x-1)^{2}+Bx+Dx(x-1)=A(x^{2}-2x+1)+Bx+D(x^{2}-x)=x^{2}(A+D)+x(-2A+B-D)+A.}
A
+
D
=
2
,
{\displaystyle A+D=2,}
−
2
A
+
B
−
D
=
−
3
,
{\displaystyle -2A+B-D=-3,}
A
=
3.
{\displaystyle A=3.}
A
=
3
;
{\displaystyle A=3;}
B
=
2
;
{\displaystyle B=2;}
D
=
−
1.
{\displaystyle D=-1.}
∫
2
x
2
−
3
x
+
3
x
3
−
2
x
2
+
x
d
x
=
3
∫
d
x
x
+
2
∫
d
x
(
x
−
1
)
2
−
∫
d
x
x
−
1
=
3
ln
|
x
|
−
2
x
−
1
−
ln
|
x
−
1
|
+
C
.
{\displaystyle \int {\frac {2x^{2}-3x+3}{x^{3}-2x^{2}+x}}dx=3\int {\frac {dx}{x}}+2\int {\frac {dx}{(x-1)^{2}}}-\int {\frac {dx}{x-1}}=3\ln |x|-{\frac {2}{x-1}}-\ln |x-1|+C.}
∫
2
x
−
1
x
2
−
5
x
+
6
d
x
=
∫
2
x
−
1
(
x
−
3
)
(
x
−
2
)
d
x
=
∫
(
A
x
−
3
+
B
x
−
2
)
d
x
,
{\displaystyle \int {2x-1 \over x^{2}-5x+6}dx=\int {2x-1 \over (x-3)(x-2)}dx=\int ({A \over x-3}+{B \over x-2})dx,}
kur
2
x
−
1
=
A
(
x
−
2
)
+
B
(
x
−
3
)
=
(
A
+
B
)
x
−
2
A
−
3
B
.
{\displaystyle 2x-1=A(x-2)+B(x-3)=(A+B)x-2A-3B.}
A
+
B
=
2
;
{\displaystyle A+B=2;}
−
2
A
−
3
B
=
−
1.
{\displaystyle -2A-3B=-1.}
Iš kur
A
=
5
;
{\displaystyle A=5;}
B
=
−
3.
{\displaystyle B=-3.}
∫
2
x
−
1
x
2
−
5
x
+
6
d
x
=
∫
(
5
x
−
3
−
3
x
−
2
)
d
x
=
5
∫
d
(
x
−
3
)
x
−
3
−
3
∫
d
(
x
−
2
)
x
−
2
=
{\displaystyle \int {2x-1 \over x^{2}-5x+6}dx=\int ({5 \over x-3}-{3 \over x-2})dx=5\int {d(x-3) \over x-3}-3\int {d(x-2) \over x-2}=}
=
5
ln
|
x
−
3
|
−
3
ln
|
x
−
2
|
+
C
=
ln
|
(
x
−
3
)
5
(
x
−
2
)
3
|
+
C
.
{\displaystyle =5\ln |x-3|-3\ln |x-2|+C=\ln |{(x-3)^{5} \over (x-2)^{3}}|+C.}
∫
9
x
3
−
30
x
2
+
28
x
−
88
(
x
2
−
6
x
+
8
)
(
x
2
+
4
)
d
x
=
∫
9
x
3
−
30
x
2
+
28
x
−
88
(
x
−
2
)
(
x
−
4
)
(
x
2
+
4
)
d
x
=
∫
(
A
x
−
2
+
B
x
−
4
+
C
x
+
D
x
2
+
4
)
d
x
;
{\displaystyle \int {9x^{3}-30x^{2}+28x-88 \over (x^{2}-6x+8)(x^{2}+4)}dx=\int {9x^{3}-30x^{2}+28x-88 \over (x-2)(x-4)(x^{2}+4)}dx=\int ({A \over x-2}+{B \over x-4}+{Cx+D \over x^{2}+4})dx;}
9
x
3
−
30
x
2
+
28
x
−
88
=
A
(
x
−
4
)
(
x
2
+
4
)
+
B
(
x
−
2
)
(
x
2
+
4
)
+
(
C
x
+
D
)
(
x
2
−
6
x
+
8
)
=
{\displaystyle 9x^{3}-30x^{2}+28x-88=A(x-4)(x^{2}+4)+B(x-2)(x^{2}+4)+(Cx+D)(x^{2}-6x+8)=}
=
(
A
+
B
+
C
)
x
3
+
(
−
4
A
−
2
B
−
6
C
+
D
)
x
2
+
(
4
A
+
4
B
+
8
C
−
6
D
)
x
+
(
−
16
A
−
8
B
+
8
D
)
;
{\displaystyle =(A+B+C)x^{3}+(-4A-2B-6C+D)x^{2}+(4A+4B+8C-6D)x+(-16A-8B+8D);}
Palyginam koeficientus prie vienodų laipsnių x .
x
3
{\displaystyle x^{3}}
A
+
B
+
C
=
9
,
{\displaystyle A+B+C=9,}
x
2
{\displaystyle x^{2}}
−
4
A
−
2
B
−
6
C
+
D
=
−
30
,
{\displaystyle -4A-2B-6C+D=-30,}
x
1
{\displaystyle x^{1}}
4
A
+
4
B
+
8
C
−
6
D
=
28
,
{\displaystyle 4A+4B+8C-6D=28,}
x
0
{\displaystyle x^{0}}
−
16
A
−
8
B
+
8
D
=
−
88
,
{\displaystyle -16A-8B+8D=-88,}
Išsprendę sistemą , randame:
A
=
5
;
{\displaystyle A=5;}
B
=
3
;
{\displaystyle B=3;}
C
=
1
;
{\displaystyle C=1;}
D
=
2.
{\displaystyle D=2.}
∫
9
x
3
−
30
x
2
+
28
x
−
88
(
x
2
−
6
x
+
8
)
(
x
2
+
4
)
d
x
=
∫
(
5
x
−
2
+
3
x
−
4
+
x
+
2
x
2
+
4
)
d
x
=
5
∫
d
(
x
−
2
)
x
−
2
+
3
∫
d
(
x
−
4
)
x
−
4
+
{\displaystyle \int {9x^{3}-30x^{2}+28x-88 \over (x^{2}-6x+8)(x^{2}+4)}dx=\int ({5 \over x-2}+{3 \over x-4}+{x+2 \over x^{2}+4})dx=5\int {d(x-2) \over x-2}+3\int {d(x-4) \over x-4}+}
+
1
2
∫
d
(
x
2
+
4
)
x
2
+
4
+
∫
2
d
x
x
2
+
4
=
5
ln
|
x
−
2
|
+
3
ln
|
x
−
4
|
+
1
2
ln
|
x
2
+
4
|
+
2
2
arctan
x
2
+
C
=
{\displaystyle +{1 \over 2}\int {d(x^{2}+4) \over x^{2}+4}+\int {2\;dx \over x^{2}+4}=5\ln |x-2|+3\ln |x-4|+{1 \over 2}\ln |x^{2}+4|+{2 \over 2}\arctan {x \over 2}+C=}
=
ln
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x
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5
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3
x
2
+
4
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+
arctan
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2
+
C
,
{\displaystyle =\ln |(x-2)^{5}(x-4)^{3}{\sqrt {x^{2}+4}}|+\arctan {x \over 2}+C,}
d
(
x
2
+
4
)
=
2
x
d
x
.
{\displaystyle d(x^{2}+4)=2xdx.}
∫
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x
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+
2
)
d
x
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x
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1
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3
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2
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=
∫
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A
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3
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A
1
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1
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2
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A
2
x
+
1
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B
x
−
2
)
d
x
.
{\displaystyle \int {(x^{2}+2)dx \over (x+1)^{3}(x-2)}=\int ({A \over (x+1)^{3}}+{A_{1} \over (x+1)^{2}}+{A_{2} \over x+1}+{B \over x-2})dx.}
x
2
+
2
=
A
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+
A
1
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1
)
(
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A
2
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2
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+
B
(
x
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1
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3
=
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A
2
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B
)
x
3
+
(
A
1
+
3
B
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x
2
+
(
A
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A
1
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3
A
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+
3
B
)
x
+
(
−
2
A
−
2
A
1
−
2
A
2
+
B
)
.
{\displaystyle x^{2}+2=A(x-2)+A_{1}(x+1)(x-2)+A_{2}(x+1)^{2}(x-2)+B(x+1)^{3}=(A_{2}+B)x^{3}+(A_{1}+3B)x^{2}+(A-A_{1}-3A_{2}+3B)x+(-2A-2A_{1}-2A_{2}+B).}
x
3
{\displaystyle x^{3}}
0
=
A
2
+
B
,
{\displaystyle 0=A_{2}+B,}
x
2
{\displaystyle x^{2}}
1
=
A
1
+
3
B
,
{\displaystyle 1=A_{1}+3B,}
x
1
{\displaystyle x^{1}}
0
=
A
−
A
1
−
3
A
2
+
3
B
,
{\displaystyle 0=A-A_{1}-3A_{2}+3B,}
x
0
{\displaystyle x^{0}}
2
=
−
2
A
−
2
A
1
−
2
A
2
+
B
.
{\displaystyle 2=-2A-2A_{1}-2A_{2}+B.}
Išsprendę sistema, randame:
A
=
−
1
;
A
1
=
1
3
;
A
2
=
−
2
9
;
B
=
2
9
.
{\displaystyle A=-1;\;A_{1}={1 \over 3};\;A_{2}=-{2 \over 9};\;B={2 \over 9}.}
∫
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2
+
2
)
d
x
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1
)
3
(
x
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2
)
=
∫
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1
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3
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3
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2
−
2
9
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9
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)
d
x
=
{\displaystyle \int {(x^{2}+2)dx \over (x+1)^{3}(x-2)}=\int (-{1 \over (x+1)^{3}}+{1 \over 3(x+1)^{2}}-{2 \over 9(x+1)}+{2 \over 9(x-2)})dx=}
=
1
2
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2
−
1
3
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x
+
1
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−
2
9
ln
|
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1
|
+
2
9
ln
|
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2
|
+
C
=
1
−
2
x
6
(
x
+
1
)
2
+
ln
|
(
x
−
2
x
+
1
)
2
9
|
+
C
.
{\displaystyle ={1 \over 2(x+1)^{2}}-{1 \over 3(x+1)}-{2 \over 9}\ln |x+1|+{2 \over 9}\ln |x-2|+C={1-2x \over 6(x+1)^{2}}+\ln |({x-2 \over x+1})^{2 \over 9}|+C.}
![{\displaystyle ax^{2}+bx+c=a(x^{2}+{b \over a}x+{c \over a})=a[x^{2}+2{b \over 2a}x+({b \over 2a})^{2}+{c \over a}-({b \over 2a})^{2}]=a[(x+{b \over 2a})^{2}+({c \over a}-{b^{2} \over 4a^{2}})].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35d14953987cc3fa7df42e4bb6ad2d1b760e1545)
![{\displaystyle \int {dx \over 2x^{2}+8x+20}={1 \over 2}\int {dx \over x^{2}+4x+10}={1 \over 2}\int {dx \over 1[(x+{4 \over 2\cdot 1})^{2}+({10 \over 1}-{4^{2} \over 4\cdot 1^{2}})]}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c75a20575a4c7bdfc11d498496755c37d395d43)
![{\displaystyle Q'(x)=[(x-1)^{3}(x-2)]'=(x^{4}-5x^{3}+9x^{2}-7x+2)',}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a70536c2e4bc3d7d6ee60d9c5ca6c5d23e93c7f1)
![{\displaystyle Q'(x)=[3(x-1)^{2}(x-2)+(x-1)^{3}]=(4x^{3}-15x^{2}+18x-7);}](https://wikimedia.org/api/rest_v1/media/math/render/svg/082e4869363a4f3c6b9d9fc45d8fe51f9ff017a6)
![{\displaystyle Q''(x)=[3(x-1)^{2}(x-2)+(x-1)^{3}]'=(4x^{3}-15x^{2}+18x-7)',}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb1551c0f2c53ca9e4e81ce5f7cfbf010e161ad1)
![{\displaystyle Q''(x)=[6(x-1)(x-2)+3(x-1)^{2}(x-2)+3(x-1)^{2}]=(12x^{2}-30x+18);}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fda58ac1aabfb5adaa8aa300803334283b57442b)
![{\displaystyle Q'''(x)=[6(x-1)(x-2)+3(x-1)^{2}(x-2)+3(x-1)^{2}]'=(12x^{2}-30x+18)',}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5de8b3a16bae924b4083d6b0f1c22276754806e)
![{\displaystyle Q'''(x)=[6(x-2)+6(x-1)+6(x-1)(x-2)+3(x-1)^{2}+6(x-1)]=(24x-30);}](https://wikimedia.org/api/rest_v1/media/math/render/svg/451bcaab58c64a47c47c381b8ec7f8ec5e5bda38)
![{\displaystyle Q'''(1)=[6(1-2)+6(1-1)+6(1-1)(x-2)+3(1-1)^{2}+6(1-1)]=-6;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52533e8776d78f2c6cdcd3f3f5ba005ccbab8140)
![{\displaystyle [u(x)\cdot v(x)]'=u'(x)v(x)+u(x)v'(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10f42bfffb76149d672b5c5cddee7271832f0dba)
![{\displaystyle [u(x)\cdot v(x)]'=u'(x)v(x)+u(x)v'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f9493ada10e8d0d5ce26f25e6b56ea568938526)
![{\displaystyle Q'(x)=[uvw]'=(uv)'w+uvw'=(u'v+uv')w+uvw'=u'vw+uv'w+uvw'.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49f03dadbe7e8d896900162488fa3528edce430b)
![{\displaystyle Q'(x)=[u_{1}u_{2}u_{3}...u_{m}]'=u_{1}'u_{2}u_{3}...u_{m}+u_{1}u_{2}'u_{3}...u_{m}+u_{1}u_{2}u_{3}'...u_{m}+...+u_{1}u_{2}u_{3}...u_{m}'.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bcd8e868587015ca0638604a02d543bd18a7603)
![{\displaystyle \int {\frac {x^{5}+x^{4}-8}{x^{3}-4x}}dx=\int [x^{2}+x+4+{\frac {4x^{2}+16x-8}{x(x+2)(x-2)}}]dx={\frac {x^{3}}{3}}+{\frac {x^{2}}{2}}+4x+4\int {\frac {x^{2}+4x-2}{x(x+2)(x-2)}}dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cbd0340be77a0fe3e78345bfffef02695e793e5)
![{\displaystyle \int {\frac {15x^{2}-4x-81}{x^{3}-13x+12}}dx=\int {\frac {15x^{2}-4x-81}{(x-3)(x+4)(x-1)}}dx=\int [{\frac {A}{x-3}}+{\frac {B}{x+4}}+{\frac {D}{x-1}}]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3687c874f19bbe25f2077a61cef854a4a39df677)
![{\displaystyle \int {\frac {x^{4}-3x^{2}-3x-2}{x^{3}-x^{2}-2x}}dx=\int [x+1-{\frac {x+2}{x(x^{2}-x-2)}}]dx;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3568a4684128cb23321341369acef400e02e9f8)
![{\displaystyle \int {\frac {2x^{2}-3x+3}{x^{3}-2x^{2}+x}}dx=\int [{\frac {A}{x}}+{\frac {B}{(x-1)^{2}}}+{\frac {D}{x-1}}]dx.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/043f92d4eb2c07a3cdf401fd84b061fd48695b3e)
