I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
=
1
2
arctan
x
2
+
x
+
1
2
(
1
−
x
)
2
+
1
2
6
ln
|
x
2
+
x
+
1
(
x
+
1
)
2
+
2
3
x
2
+
x
+
1
(
x
+
1
)
2
−
2
3
|
+
C
.
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {x^{2}+x+1}{2(1-x)^{2}}}}+{\frac {1}{2{\sqrt {6}}}}\ln {\Big |}{\frac {{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}-{\sqrt {\frac {2}{3}}}}}{\Big |}+C.}
Arba
I
=
I
1
+
I
2
.
{\displaystyle I=I_{1}+I_{2}.}
I
1
,
{\displaystyle I_{1},}
x kinta nuo 0 iki 3:
I
1
=
1
2
arctan
x
2
+
x
+
1
2
(
1
−
x
)
2
|
0
3
=
1
2
arctan
3
2
+
3
+
1
2
(
1
−
3
)
2
−
1
2
arctan
0
2
+
0
+
1
2
(
1
−
0
)
2
=
1
2
arctan
13
8
−
1
2
arctan
1
2
=
{\displaystyle I_{1}={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {x^{2}+x+1}{2(1-x)^{2}}}}|_{0}^{3}={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {3^{2}+3+1}{2(1-3)^{2}}}}-{\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {0^{2}+0+1}{2(1-0)^{2}}}}={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {13}{8}}}-{\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {1}{2}}}=}
=
1
2
arctan
1.625
−
1
2
arctan
0.5
=
1
2
(
arctan
(
1.274754878
)
−
arctan
(
0.70710678
)
)
=
1
2
(
0.905600271782
−
0.61547970867
)
=
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\sqrt {1.625}}-{\frac {1}{\sqrt {2}}}\arctan {\sqrt {0.5}}={\frac {1}{\sqrt {2}}}(\arctan(1.274754878)-\arctan(0.70710678))={\frac {1}{\sqrt {2}}}(0.905600271782-0.61547970867)=}
=0.29012056311169063853376698805293/1.4142135623730950488016887242097=0.20514621753793618370771135085263.
Toliau apskaičiuosime integralą
I
2
,
{\displaystyle I_{2},}
x kinta nuo 0 iki 3:
I
2
=
1
2
6
ln
x
2
+
x
+
1
(
x
+
1
)
2
+
2
3
x
2
+
x
+
1
(
x
+
1
)
2
−
2
3
|
0
3
=
{\displaystyle I_{2}={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {x^{2}+x+1}{(x+1)^{2}}}}-{\sqrt {\frac {2}{3}}}}}{\Big |}_{0}^{3}=}
=
1
2
6
ln
3
2
+
3
+
1
(
3
+
1
)
2
+
2
3
3
2
+
3
+
1
(
3
+
1
)
2
−
2
3
−
1
2
6
ln
0
2
+
0
+
1
(
0
+
1
)
2
+
2
3
0
2
+
0
+
1
(
0
+
1
)
2
−
2
3
=
1
2
6
ln
13
16
+
2
3
13
16
−
2
3
−
1
2
6
ln
1
1
+
2
3
1
1
−
2
3
=
{\displaystyle ={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {3^{2}+3+1}{(3+1)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {3^{2}+3+1}{(3+1)^{2}}}}-{\sqrt {\frac {2}{3}}}}}-{\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {0^{2}+0+1}{(0+1)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {0^{2}+0+1}{(0+1)^{2}}}}-{\sqrt {\frac {2}{3}}}}}={\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {13}{16}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {13}{16}}}-{\sqrt {\frac {2}{3}}}}}-{\frac {1}{2{\sqrt {6}}}}\ln {\frac {{\sqrt {\frac {1}{1}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {1}{1}}}-{\sqrt {\frac {2}{3}}}}}=}
=
1
2
6
(
ln
0.8125
+
0.
(
6
)
0.8125
−
0.
(
6
)
−
ln
1
+
0.
(
6
)
1
−
0.
(
6
)
)
=
{\displaystyle ={\frac {1}{2{\sqrt {6}}}}{\Big (}\ln {\frac {{\sqrt {0.8125}}+{\sqrt {0.(6)}}}{{\sqrt {0.8125}}-{\sqrt {0.(6)}}}}-\ln {\frac {1+{\sqrt {0.(6)}}}{1-{\sqrt {0.(6)}}}}{\Big )}=}
=0.20412414523193150818310700622549 * [ln(1.7178843997937233560122333417696/0.08489123793827129054737729196566) - ln(1.816496580927726032732428024902/0.18350341907227396726757197509804)]=
=0.20412414523193150818310700622549 * [ln(20.236298132946110691158905973668) - ln(9.8989794855663561963945681494118)]=
=0.20412414523193150818310700622549 * [3.0074779291214254623297105529653 - 2.292431669561177687800787311348]=
=0.20412414523193150818310700622549 * 0.7150462595602477745289232416173 = 0.14595820653402541036436689756671.
Taigi, sudėjus
I
1
{\displaystyle I_{1}}
I
2
{\displaystyle I_{2}}
I = 0.20514621753793618370771135085263 + 0.14595820653402541036436689756671 = 0.35110442407196159407207824841934.
Toliau palyginsime gautą integralo I atsakymą su funkcijos
f
(
x
)
=
1
(
x
2
−
x
+
1
)
x
2
+
x
+
1
{\displaystyle f(x)={\frac {1}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}}
I reikšmę. Taigi,
S
m
a
x
=
3
⋅
f
(
0
)
=
3
⋅
1
(
0
2
−
0
+
1
)
0
2
+
0
+
1
=
3.
{\displaystyle S_{max}=3\cdot f(0)=3\cdot {\frac {1}{(0^{2}-0+1){\sqrt {0^{2}+0+1}}}}=3.}
S
m
i
n
=
3
⋅
f
(
3
)
=
3
⋅
1
(
3
2
−
3
+
1
)
3
2
+
3
+
1
=
3
⋅
1
7
13
=
3
7
⋅
3.60555127546
=
{\displaystyle S_{min}=3\cdot f(3)=3\cdot {\frac {1}{(3^{2}-3+1){\sqrt {3^{2}+3+1}}}}=3\cdot {\frac {1}{7{\sqrt {13}}}}={\frac {3}{7\cdot 3.60555127546}}=}
= 3/25.238858928247925051834548872293 = 0.11886432776254909757535894288364.
Matome, kad
S
m
i
n
<
I
<
S
m
a
x
{\displaystyle S_{min}<I<S_{max}\;}
0.11886432776254909757535894288364 < 0.35110442407196159407207824841934 < 3.
Vadinasi, integralo I skaičiavime didelių klaidų nėra ir gautas atsakymas (
I
≈
0.35110442407196
{\displaystyle I\approx 0.35110442407196}
Toks Free Pascal kodas: var a:longint; b,c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.000000003/((sqr(a*0.000000003)-a*0.000000003+1)*sqrt(sqr(a*0.000000003)+a*0.000000003+1));
writeln(c);
readln;
end.
duoda atsakymą "1.2918337052676980E+000" po 21 sekundės ant 4.16 GHz dažniu veikiančio procesoriaus (patį pirmą kartą paleidus šį kodą, gaunamas atsakymas po 43 sekundžių). Tai reiškia, kad plotas po funkcija
f
(
x
)
=
1
(
x
2
−
x
+
1
)
x
2
+
x
+
1
{\displaystyle f(x)={\frac {1}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}}
S
=
I
=
1.2918337052676980.
{\displaystyle S=I=1.2918337052676980.}
Kaip matome su Free Pascal skaičiuotas plotas skiriasi nuo gauto atsakymo (ploto) integravimo budu. Bandžiau įstatyti į integralą
I
2
{\displaystyle I_{2}\;}
8
3
{\displaystyle {\sqrt {\frac {8}{3}}}}
2
3
,
{\displaystyle {\sqrt {\frac {2}{3}}},}
I reikšmė (integralas
I
2
{\displaystyle I_{2}}
I
2
{\displaystyle I_{2}}
Internetinis integratorius https://www.wolframalpha.com/calculators/integral-calculator/ šitą integralą integruoja taip:
∫
1
(
x
2
−
x
+
1
)
x
2
+
x
+
1
d
x
=
1
4
3
(
1
−
i
3
(
3
+
−
i
)
tanh
−
1
(
(
4
−
2
i
3
)
x
−
i
3
+
5
4
1
−
i
3
x
2
+
x
+
1
)
+
1
+
i
3
(
3
+
i
)
tanh
−
1
(
(
4
+
2
i
3
)
x
+
i
3
+
5
4
1
+
i
3
x
2
+
x
+
1
)
)
+
C
.
{\displaystyle \int {\frac {1}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}dx={\frac {1}{4{\sqrt {3}}}}{\Big (}{\sqrt {1-i{\sqrt {3}}}}({\sqrt {3}}+-i)\tanh ^{-1}{\Big (}{\frac {(4-2i{\sqrt {3}})x-i{\sqrt {3}}+5}{4{\sqrt {1-i{\sqrt {3}}}}{\sqrt {x^{2}+x+1}}}}{\Big )}+{\sqrt {1+i{\sqrt {3}}}}({\sqrt {3}}+i)\tanh ^{-1}{\Big (}{\frac {(4+2i{\sqrt {3}})x+i{\sqrt {3}}+5}{4{\sqrt {1+i{\sqrt {3}}}}{\sqrt {x^{2}+x+1}}}}{\Big )}{\Big )}+C.}
tanh
−
1
(
x
)
{\displaystyle \tanh ^{-1}(x)}
Sprendžiant iš tokio neteisingo integravimo, nenustebčiau jei dalis sudetingų integralų iš integralų lentelių yra neteisingi arba neduoda ploto po funkciją, kurią jie integruoja. Gal paėmus išvestinę gauto integralo, tarkim šitame pavyzdyje, gausime pradinę funkciją
f
(
x
)
=
1
(
x
2
−
x
+
1
)
x
2
+
x
+
1
,
{\displaystyle f(x)={\frac {1}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}},}
Kad integravimo keliu gautas atsakymas
I
≈
0.35110442407196
{\displaystyle I\approx 0.35110442407196}
S
1
=
1
⋅
f
(
1
)
=
1
⋅
1
(
1
2
−
1
+
1
)
1
2
+
1
+
1
=
1
⋅
1
3
=
{\displaystyle S_{1}=1\cdot f(1)=1\cdot {\frac {1}{(1^{2}-1+1){\sqrt {1^{2}+1+1}}}}=1\cdot {\frac {1}{\sqrt {3}}}=}
=0.57735026918962576450914878050196.
Su reikšmėm mažesnėm nei x=1 funkcija
f
(
x
)
=
1
(
x
2
−
x
+
1
)
x
2
+
x
+
1
{\displaystyle f(x)={\frac {1}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}}
S
1
{\displaystyle S_{1}}
0.57735026918962576450914878050196 > 0.35110442407196159407207824841934
ir tai įrodo, kad integravimo budu (nuo 0 iki 3) gautas atsakymas (plotas) yra neteisingas. Paraboloid (aptarimas ) 17:24, 26 gegužės 2023 (UTC) Atsakyti
Apskaičiuosime pirmojo tipo integralo pavyzdžio reikšmę, kai x kinta nuo 0 iki 2:
∫
0
2
x
3
1
+
2
x
−
x
2
d
x
=
(
(
−
19
6
−
5
6
x
−
1
3
x
2
)
1
+
2
x
−
x
2
+
4
arcsin
x
−
1
2
)
|
0
2
=
{\displaystyle \int _{0}^{2}{\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}\;dx={\Big (}{\Big (}-{\frac {19}{6}}-{\frac {5}{6}}x-{\frac {1}{3}}x^{2}{\Big )}{\sqrt {1+2x-x^{2}}}+4\arcsin {\frac {x-1}{\sqrt {2}}}{\Big )}|_{0}^{2}=}
=
(
(
−
19
6
−
5
6
⋅
2
−
1
3
⋅
2
2
)
1
+
2
⋅
2
−
2
2
+
4
arcsin
2
−
1
2
)
−
(
(
−
19
6
−
5
6
⋅
0
−
1
3
⋅
0
2
)
1
+
2
⋅
0
−
0
2
+
4
arcsin
0
−
1
2
)
=
{\displaystyle ={\Big (}{\Big (}-{\frac {19}{6}}-{\frac {5}{6}}\cdot 2-{\frac {1}{3}}\cdot 2^{2}{\Big )}{\sqrt {1+2\cdot 2-2^{2}}}+4\arcsin {\frac {2-1}{\sqrt {2}}}{\Big )}-{\Big (}{\Big (}-{\frac {19}{6}}-{\frac {5}{6}}\cdot 0-{\frac {1}{3}}\cdot 0^{2}{\Big )}{\sqrt {1+2\cdot 0-0^{2}}}+4\arcsin {\frac {0-1}{\sqrt {2}}}{\Big )}=}
=
(
(
−
19
6
−
10
6
−
4
3
)
1
+
4
arcsin
1
2
)
−
(
−
19
6
1
+
4
arcsin
−
1
2
)
=
{\displaystyle ={\Big (}{\Big (}-{\frac {19}{6}}-{\frac {10}{6}}-{\frac {4}{3}}{\Big )}{\sqrt {1}}+4\arcsin {\frac {1}{\sqrt {2}}}{\Big )}-{\Big (}-{\frac {19}{6}}{\sqrt {1}}+4\arcsin {\frac {-1}{\sqrt {2}}}{\Big )}=}
=
(
−
19
+
10
+
8
6
+
4
π
4
)
−
(
−
19
6
+
4
⋅
−
π
4
)
=
{\displaystyle ={\Big (}-{\frac {19+10+8}{6}}+4{\frac {\pi }{4}}{\Big )}-{\Big (}-{\frac {19}{6}}+4\cdot {\frac {-\pi }{4}}{\Big )}=}
=
−
37
6
+
π
+
19
6
+
π
=
19
−
37
6
+
2
π
=
−
18
6
+
2
π
=
2
π
−
3
=
{\displaystyle =-{\frac {37}{6}}+\pi +{\frac {19}{6}}+\pi ={\frac {19-37}{6}}+2\pi =-{\frac {18}{6}}+2\pi =2\pi -3=}
= 6.28318530717958647692528676655 - 3 = 3.283185307179586476925286766559. Patikriname ar gautas atsakymas nėra didesnis už maksimalią funkcijos
f
(
x
)
=
x
3
1
+
2
x
−
x
2
{\displaystyle f(x)={\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}}
S
max
=
2
f
(
2
)
=
2
⋅
2
3
1
+
2
⋅
2
−
2
2
=
2
⋅
8
1
=
16
;
{\displaystyle S_{\text{max}}=2f(2)=2\cdot {\frac {2^{3}}{\sqrt {1+2\cdot 2-2^{2}}}}=2\cdot {\frac {8}{\sqrt {1}}}=16;}
S
min
=
2
f
(
0
)
=
2
⋅
0
3
1
+
2
⋅
0
−
0
2
=
2
⋅
0
1
=
0.
{\displaystyle S_{\text{min}}=2f(0)=2\cdot {\frac {0^{3}}{\sqrt {1+2\cdot 0-0^{2}}}}=2\cdot {\frac {0}{\sqrt {1}}}=0.}
Matome, kad gautas integralo atsakymas ~3.2831853 tenkina nusakytas sąlygas, todėl turi visus šansus buti teisingu, nes
0 < 3.2831853 < 16. Toks Free Pascal kodas: var a:longint; b,c:real;
begin
for a:=1 to 1000000000 do
c:=c+a*0.000000002*sqr(a*0.000000002)/sqrt(1+a*0.000000004-sqr(a*0.000000002));
writeln(0.000000002*c);
readln;
end.
duoda rezultatą "3.28318531517779016492E+0000" po 21 sekundės su 4.16 GHz procesorium (pirmą kartą [paleidus] šitas kodas duodą šitą rezultatą po 43 sekundžių). Gavome pirmus 8 teisingus skaitmenis, kai plotą po funkciją
f
(
x
)
=
x
3
1
+
2
x
−
x
2
{\displaystyle f(x)={\frac {x^{3}}{\sqrt {1+2x-x^{2}}}}}
Apskaičiuosime pavyzdyje gautą integralo reikšmę, kai x kinta nuo 0 iki 0.8:
I
=
∫
0
0.8
1
+
x
1
−
x
⋅
d
x
1
−
x
=
(
2
1
+
x
1
−
x
−
2
arctan
1
+
x
1
−
x
)
|
0
0.8
=
{\displaystyle I=\int _{0}^{0.8}{\sqrt {\frac {1+x}{1-x}}}\cdot {\frac {dx}{1-x}}={\Big (}2{\sqrt {\frac {1+x}{1-x}}}-2\arctan {\sqrt {\frac {1+x}{1-x}}}{\Big )}{\Big |}_{0}^{0.8}=}
=
(
2
1
+
0.8
1
−
0.8
−
2
arctan
1
+
0.8
1
−
0.8
)
−
(
2
1
+
0
1
−
0
−
2
arctan
1
+
0
1
−
0
)
=
{\displaystyle ={\Big (}2{\sqrt {\frac {1+0.8}{1-0.8}}}-2\arctan {\sqrt {\frac {1+0.8}{1-0.8}}}{\Big )}-{\Big (}2{\sqrt {\frac {1+0}{1-0}}}-2\arctan {\sqrt {\frac {1+0}{1-0}}}{\Big )}=}
=
(
2
1.8
0.2
−
2
arctan
1.8
0.2
)
−
(
2
1
1
−
2
arctan
1
1
)
=
{\displaystyle ={\Big (}2{\sqrt {\frac {1.8}{0.2}}}-2\arctan {\sqrt {\frac {1.8}{0.2}}}{\Big )}-{\Big (}2{\sqrt {\frac {1}{1}}}-2\arctan {\sqrt {\frac {1}{1}}}{\Big )}=}
=
(
2
9
−
2
arctan
9
)
−
(
2
1
−
2
arctan
1
)
=
{\displaystyle ={\Big (}2{\sqrt {9}}-2\arctan {\sqrt {9}}{\Big )}-{\Big (}2{\sqrt {1}}-2\arctan {\sqrt {1}}{\Big )}=}
=
(
2
⋅
3
−
2
arctan
3
)
−
(
2
−
2
arctan
1
)
=
{\displaystyle ={\Big (}2\cdot 3-2\arctan 3{\Big )}-{\Big (}2-2\arctan 1{\Big )}=}
=
6
−
2
arctan
3
−
2
+
2
arctan
1
=
4
−
2
arctan
3
+
2
arctan
1
=
{\displaystyle =6-2\arctan 3-2+2\arctan 1=4-2\arctan 3+2\arctan 1=}
= 4 - 2*1.2490457723982544258299170772811 + 2*0.78539816339744830961566084581988 =
= 4 - 2.4980915447965088516598341545622 + 1.5707963267948966192313216916398 =
= 3.0727047819983877675714875370776. Apskaičiuosime kokių reikšmių šito pavyzdžio integralas viršiti negali (funkcijai
f
(
x
)
=
1
+
x
1
−
x
⋅
1
1
−
x
{\displaystyle f(x)={\sqrt {\frac {1+x}{1-x}}}\cdot {\frac {1}{1-x}}}
S
max
=
0.8
⋅
f
(
0.8
)
=
0.8
1
+
0.8
1
−
0.8
⋅
1
1
−
0.8
=
0.8
1.8
0.2
⋅
1
0.2
=
{\displaystyle S_{\text{max}}=0.8\cdot f(0.8)=0.8{\sqrt {\frac {1+0.8}{1-0.8}}}\cdot {\frac {1}{1-0.8}}=0.8{\sqrt {\frac {1.8}{0.2}}}\cdot {\frac {1}{0.2}}=}
=
0.8
9
⋅
5
=
0.8
⋅
3
⋅
5
=
0.8
⋅
15
=
12.
{\displaystyle =0.8{\sqrt {9}}\cdot 5=0.8\cdot 3\cdot 5=0.8\cdot 15=12.}
S
min
=
0.8
⋅
f
(
0
)
=
0.8
1
+
0
1
−
0
⋅
1
1
−
0
=
0.8
1
⋅
1
=
0.8.
{\displaystyle S_{\text{min}}=0.8\cdot f(0)=0.8{\sqrt {\frac {1+0}{1-0}}}\cdot {\frac {1}{1-0}}=0.8{\sqrt {1}}\cdot 1=0.8.}
Matome, kad integravimo atsakymas ~3.07270478 gali būti teisingas, nes nelygybė
S
max
>
I
>
S
min
,
{\displaystyle S_{\text{max}}>I>S_{\text{min}},}
12 > 3.0727047819983877675714875370776 > 0.8
tenkinama. Toks Free Pascal kodas skaičiuoja plotą po funkcijos
f
(
x
)
=
1
+
x
1
−
x
⋅
1
1
−
x
{\displaystyle f(x)={\sqrt {\frac {1+x}{1-x}}}\cdot {\frac {1}{1-x}}}
var a:longint; b,c:real;
begin
for a:=1 to 1000000000 do
c:=c+0.0000000008*sqrt((1+a*0.0000000008)/(1-a*0.0000000008))/(1-a*0.0000000008);
writeln(c);
readln;
end.
ir duoda rezultatą "3.0727047875963005E+000" po 26 sekundžių su 4.16 GHz procesium (per pirmą kodo paleidimą duoda šį rezultatą po 46-47 sekundžių). Gavome pirmus 9 teisingus Free Pascal atsakymo skaitmenis (kai plotą po funkcija f(x) suskaldėme į milijardą dalių). Vadinasi, integralas I apskaičiuotas teisingai. Truputi labiau optimizuotas kodas (kuriame viena daugybos operacija mažiau): var a:longint; b,c:real;
begin
for a:=1 to 1000000000 do
c:=c+sqrt((1+a*0.0000000008)/(1-a*0.0000000008))/(1-a*0.0000000008);
writeln(0.0000000008*c);
readln;
end.
duoda rezultatą "3.07270478759553108223E+0000" po 25 sekundžių (pirmą kartą - po 46 sekundžių) su 4.16 GHz dažniu veikiančiu procesorium. Išeina, kad per ~1 sekundę atliekama milijardas daugybos operacijų. O procesorius veikia 4.16 GHz dažnių. Tai gaunasi maždaug 4.16 ciklų vienai daugybos operacijai. Bet tai netikslu, nes sekundės įvertinimas yra netikslus (gali būti, pavyzdžiui, 0.5 sekundės arba 1.5 sekundės ilgesnis skaičiavimo laikas). Įstatysime keitinį
x
=
μ
t
+
ν
1
+
t
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;}
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
a
x
2
+
b
x
+
c
=
a
(
μ
t
+
ν
1
+
t
)
2
+
b
μ
t
+
ν
1
+
t
+
c
=
{\displaystyle ax^{2}+bx+c=a{\Big (}{\frac {\mu t+\nu }{1+t}}{\Big )}^{2}+b{\frac {\mu t+\nu }{1+t}}+c=}
=
a
μ
2
t
2
+
2
μ
t
ν
+
ν
2
(
1
+
t
)
2
+
b
μ
t
+
ν
1
+
t
+
c
=
{\displaystyle =a{\frac {\mu ^{2}t^{2}+2\mu t\nu +\nu ^{2}}{(1+t)^{2}}}+b{\frac {\mu t+\nu }{1+t}}+c=}
=
a
μ
2
t
2
+
2
μ
t
ν
+
ν
2
(
1
+
t
)
2
+
b
(
μ
t
+
ν
)
(
1
+
t
)
(
1
+
t
)
2
+
c
1
+
2
t
+
t
2
(
1
+
t
)
2
=
{\displaystyle =a{\frac {\mu ^{2}t^{2}+2\mu t\nu +\nu ^{2}}{(1+t)^{2}}}+b{\frac {(\mu t+\nu )(1+t)}{(1+t)^{2}}}+c{\frac {1+2t+t^{2}}{(1+t)^{2}}}=}
=
a
μ
2
t
2
+
2
μ
t
ν
+
ν
2
(
1
+
t
)
2
+
b
μ
t
+
ν
+
μ
t
2
+
ν
t
(
1
+
t
)
2
+
c
1
+
2
t
+
t
2
(
1
+
t
)
2
=
{\displaystyle =a{\frac {\mu ^{2}t^{2}+2\mu t\nu +\nu ^{2}}{(1+t)^{2}}}+b{\frac {\mu t+\nu +\mu t^{2}+\nu t}{(1+t)^{2}}}+c{\frac {1+2t+t^{2}}{(1+t)^{2}}}=}
=
a
μ
2
t
2
+
2
a
μ
t
ν
+
a
ν
2
(
1
+
t
)
2
+
b
μ
t
+
b
ν
+
b
μ
t
2
+
b
ν
t
(
1
+
t
)
2
+
c
+
2
c
t
+
c
t
2
(
1
+
t
)
2
=
{\displaystyle ={\frac {a\mu ^{2}t^{2}+2a\mu t\nu +a\nu ^{2}}{(1+t)^{2}}}+{\frac {b\mu t+b\nu +b\mu t^{2}+b\nu t}{(1+t)^{2}}}+{\frac {c+2ct+ct^{2}}{(1+t)^{2}}}=}
=
a
μ
2
t
2
+
2
a
μ
t
ν
+
a
ν
2
+
b
μ
t
+
b
ν
+
b
μ
t
2
+
b
ν
t
+
c
+
2
c
t
+
c
t
2
(
1
+
t
)
2
=
{\displaystyle ={\frac {a\mu ^{2}t^{2}+2a\mu t\nu +a\nu ^{2}+b\mu t+b\nu +b\mu t^{2}+b\nu t+c+2ct+ct^{2}}{(1+t)^{2}}}=}
=
a
μ
2
t
2
+
b
μ
t
2
+
c
t
2
+
2
a
μ
ν
t
+
b
μ
t
+
b
ν
t
+
2
c
t
+
a
ν
2
+
b
ν
+
c
(
1
+
t
)
2
=
{\displaystyle ={\frac {a\mu ^{2}t^{2}+b\mu t^{2}+ct^{2}+2a\mu \nu t+b\mu t+b\nu t+2ct+a\nu ^{2}+b\nu +c}{(1+t)^{2}}}=}
=
(
a
μ
2
+
b
μ
+
c
)
t
2
+
[
2
a
μ
ν
+
b
μ
+
b
ν
+
2
c
]
t
+
(
a
ν
2
+
b
ν
+
c
)
(
1
+
t
)
2
=
{\displaystyle ={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2a\mu \nu +b\mu +b\nu +2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}=}
=
(
a
μ
2
+
b
μ
+
c
)
t
2
+
[
2
μ
ν
a
+
b
(
μ
+
ν
)
+
2
c
]
t
+
(
a
ν
2
+
b
ν
+
c
)
(
1
+
t
)
2
.
{\displaystyle ={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2\mu \nu a+b(\mu +\nu )+2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}.}
Gavome tokią pačią išraišką kaip teorijoje. Vadinasi, viskas apskaičiuota teisingai.
Į polinomą
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
a
=
1
,
b
=
p
,
c
=
q
{\displaystyle a=1,\;b=p,\;c=q\;}
x
2
+
p
x
+
q
{\displaystyle x^{2}+px+q}
x
=
μ
t
+
ν
1
+
t
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;}
x
2
+
p
x
+
q
=
(
μ
2
+
p
μ
+
q
)
t
2
+
[
2
μ
ν
+
p
(
μ
+
ν
)
+
2
q
]
t
+
(
ν
2
+
p
ν
+
q
)
(
1
+
t
)
2
.
{\displaystyle x^{2}+px+q={\frac {(\mu ^{2}+p\mu +q)t^{2}+[2\mu \nu +p(\mu +\nu )+2q]t+(\nu ^{2}+p\nu +q)}{(1+t)^{2}}}.}
Ir dabar gauname tokią pačią polinomo
x
2
+
p
x
+
q
{\displaystyle x^{2}+px+q}
Kadangi
x
=
μ
t
+
ν
1
+
t
,
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;,}
(
μ
t
+
ν
1
+
t
)
′
=
(
μ
t
+
ν
)
′
(
1
+
t
)
−
(
μ
t
+
ν
)
(
1
+
t
)
′
(
1
+
t
)
2
=
μ
(
1
+
t
)
−
(
μ
t
+
ν
)
(
1
+
t
)
2
=
μ
+
μ
t
−
μ
t
−
ν
(
1
+
t
)
2
=
μ
−
ν
(
1
+
t
)
2
.
{\displaystyle {\Big (}{\frac {\mu t+\nu }{1+t}}{\Big )}'={\frac {(\mu t+\nu )'(1+t)-(\mu t+\nu )(1+t)'}{(1+t)^{2}}}={\frac {\mu (1+t)-(\mu t+\nu )}{(1+t)^{2}}}={\frac {\mu +\mu t-\mu t-\nu }{(1+t)^{2}}}={\frac {\mu -\nu }{(1+t)^{2}}}\,.}
Todėl
d
x
=
μ
−
ν
(
1
+
t
)
2
d
t
.
{\displaystyle dx={\frac {\mu -\nu }{(1+t)^{2}}}\,dt.}
Pavyzdys (kurio sprendimas truputi kitoks). Apskaičiuosime integralą
I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
.
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}.}
III tipo integralas. Kadangi jis netenkina (7.75) sąlygos, tai pirmiausia turime padaryti (7.76) keitinį (
x
=
μ
t
+
ν
1
+
t
(
7.76
)
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;\;(7.76)}
x
2
+
x
+
1
=
(
μ
2
+
μ
+
1
)
t
2
+
[
2
μ
ν
+
(
μ
+
ν
)
+
2
]
t
+
(
ν
2
+
ν
+
1
)
(
1
+
t
)
2
,
{\displaystyle x^{2}+x+1={\frac {(\mu ^{2}+\mu +1)t^{2}+[2\mu \nu +(\mu +\nu )+2]t+(\nu ^{2}+\nu +1)}{(1+t)^{2}}}\;,}
x
2
−
x
+
1
=
(
μ
2
−
μ
+
1
)
t
2
+
[
2
μ
ν
−
(
μ
+
ν
)
+
2
]
t
+
(
ν
2
−
ν
+
1
)
(
1
+
t
)
2
.
{\displaystyle x^{2}-x+1={\frac {(\mu ^{2}-\mu +1)t^{2}+[2\mu \nu -(\mu +\nu )+2]t+(\nu ^{2}-\nu +1)}{(1+t)^{2}}}\;.}
Koeficientus
μ
{\displaystyle \mu }
ν
{\displaystyle \nu }
{
2
μ
ν
+
(
μ
+
ν
)
+
2
=
0
,
2
μ
ν
−
(
μ
+
ν
)
+
2
=
0.
{\displaystyle {\begin{cases}2\mu \nu +(\mu +\nu )+2=0,&\\2\mu \nu -(\mu +\nu )+2=0.&\end{cases}}}
Lengva įsitikinti, kad
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1\;}
μ
=
−
1
,
ν
=
1
{\displaystyle \mu =-1,\;\nu =1}
μ
=
−
1
,
ν
=
1.
{\displaystyle \mu =-1,\;\nu =1.}
x
=
μ
t
+
ν
1
+
t
=
−
t
+
1
t
+
1
,
{\displaystyle x={\frac {\mu t+\nu }{1+t}}={\frac {-t+1}{t+1}}\;,}
x
(
t
+
1
)
=
1
−
t
,
x
t
+
t
=
1
−
x
,
t
(
1
+
x
)
=
1
−
x
,
{\displaystyle x(t+1)=1-t,\;xt+t=1-x,\;t(1+x)=1-x,\;}
t
=
1
−
x
1
+
x
;
{\displaystyle t={\frac {1-x}{1+x}}\;;}
iš truputi aukščiau
d
x
=
μ
−
ν
(
1
+
t
)
2
d
t
=
−
1
−
1
(
1
+
t
)
2
d
t
=
−
2
(
1
+
t
)
2
d
t
{\displaystyle dx={\frac {\mu -\nu }{(1+t)^{2}}}\,dt={\frac {-1-1}{(1+t)^{2}}}\,dt=-{\frac {2}{(1+t)^{2}}}\,dt}
arba
d
x
=
−
1
⋅
(
t
+
1
)
−
(
−
t
+
1
)
⋅
1
(
t
+
1
)
2
d
t
=
−
2
d
t
(
t
+
1
)
2
.
{\displaystyle dx={\frac {-1\cdot (t+1)-(-t+1)\cdot 1}{(t+1)^{2}}}\;dt={\frac {-2\;dt}{(t+1)^{2}}}\;.}
Toliau
x
2
+
x
+
1
=
(
−
t
+
1
)
2
(
t
+
1
)
2
+
−
t
+
1
t
+
1
+
1
=
t
2
−
2
t
+
1
(
t
+
1
)
2
+
−
t
+
1
t
+
1
+
1
=
(
t
2
−
2
t
+
1
)
+
(
t
+
1
)
(
−
t
+
1
)
+
(
t
+
1
)
2
(
t
+
1
)
2
=
{\displaystyle x^{2}+x+1={\frac {(-t+1)^{2}}{(t+1)^{2}}}+{\frac {-t+1}{t+1}}+1={\frac {t^{2}-2t+1}{(t+1)^{2}}}+{\frac {-t+1}{t+1}}+1={\frac {(t^{2}-2t+1)+(t+1)(-t+1)+(t+1)^{2}}{(t+1)^{2}}}=}
=
(
t
2
−
2
t
+
1
)
+
(
1
−
t
2
)
+
(
t
2
+
2
t
+
1
)
(
t
+
1
)
2
=
t
2
+
3
(
t
+
1
)
2
,
{\displaystyle ={\frac {(t^{2}-2t+1)+(1-t^{2})+(t^{2}+2t+1)}{(t+1)^{2}}}={\frac {t^{2}+3}{(t+1)^{2}}}\;,}
x
2
−
x
+
1
=
(
−
t
+
1
)
2
(
t
+
1
)
2
−
−
t
+
1
t
+
1
+
1
=
t
2
−
2
t
+
1
(
t
+
1
)
2
−
−
t
+
1
t
+
1
+
1
=
(
t
2
−
2
t
+
1
)
−
(
t
+
1
)
(
−
t
+
1
)
+
(
t
+
1
)
2
(
t
+
1
)
2
=
{\displaystyle x^{2}-x+1={\frac {(-t+1)^{2}}{(t+1)^{2}}}-{\frac {-t+1}{t+1}}+1={\frac {t^{2}-2t+1}{(t+1)^{2}}}-{\frac {-t+1}{t+1}}+1={\frac {(t^{2}-2t+1)-(t+1)(-t+1)+(t+1)^{2}}{(t+1)^{2}}}=}
=
(
t
2
−
2
t
+
1
)
−
(
1
−
t
2
)
+
(
t
2
+
2
t
+
1
)
(
t
+
1
)
2
=
3
t
2
+
1
(
t
+
1
)
2
.
{\displaystyle ={\frac {(t^{2}-2t+1)-(1-t^{2})+(t^{2}+2t+1)}{(t+1)^{2}}}={\frac {3t^{2}+1}{(t+1)^{2}}}\;.}
Nagrinėjamasis integralas virsta šitokiu:
I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
=
∫
1
3
t
2
+
1
(
t
+
1
)
2
t
2
+
3
(
t
+
1
)
2
⋅
−
2
d
t
(
t
+
1
)
2
=
−
2
∫
1
(
3
t
2
+
1
)
t
2
+
3
(
t
+
1
)
2
d
t
=
−
2
∫
(
t
+
1
)
d
t
(
3
t
2
+
1
)
t
2
+
3
=
I
1
+
I
2
,
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}=\int {\frac {1}{{\frac {3t^{2}+1}{(t+1)^{2}}}{\sqrt {\frac {t^{2}+3}{(t+1)^{2}}}}}}\cdot {\frac {-2\;dt}{(t+1)^{2}}}=-2\int {\frac {1}{(3t^{2}+1){\sqrt {\frac {t^{2}+3}{(t+1)^{2}}}}}}\;dt=-2\int {\frac {(t+1)dt}{(3t^{2}+1){\sqrt {t^{2}+3}}}}=I_{1}+I_{2},}
jei
I
1
=
−
2
∫
t
d
t
(
3
t
2
+
1
)
t
2
+
3
,
I
2
=
−
2
∫
d
t
(
3
t
2
+
1
)
t
2
+
3
.
{\displaystyle I_{1}=-2\int {\frac {t\;dt}{(3t^{2}+1){\sqrt {t^{2}+3}}}}\;,\quad I_{2}=-2\int {\frac {dt}{(3t^{2}+1){\sqrt {t^{2}+3}}}}.}
Integralui
I
1
{\displaystyle I_{1}}
u
=
t
2
+
3
,
{\displaystyle u={\sqrt {t^{2}+3}},\;}
I
2
{\displaystyle I_{2}}
v
=
1
+
3
t
2
.
{\displaystyle v={\sqrt {1+{\frac {3}{t^{2}}}}}.}
Tada integralui
I
1
{\displaystyle I_{1}}
u
2
=
t
2
+
3
,
u
2
−
3
=
t
2
,
t
2
=
u
2
−
3
,
{\displaystyle u^{2}=t^{2}+3,\;u^{2}-3=t^{2},\;t^{2}=u^{2}-3\;,}
d
(
t
2
)
=
d
(
u
2
−
3
)
=
2
u
d
u
{\displaystyle d(t^{2})=d(u^{2}-3)=2u\;du\;}
I
1
=
−
2
∫
t
d
t
(
3
t
2
+
1
)
t
2
+
3
=
−
2
2
∫
d
(
t
2
)
(
3
t
2
+
1
)
t
2
+
3
=
−
∫
2
u
d
u
(
3
(
u
2
−
3
)
+
1
)
⋅
u
=
−
2
∫
d
u
3
u
2
−
9
+
1
=
{\displaystyle I_{1}=-2\int {\frac {t\;dt}{(3t^{2}+1){\sqrt {t^{2}+3}}}}={\frac {-2}{2}}\int {\frac {d(t^{2})}{(3t^{2}+1){\sqrt {t^{2}+3}}}}=-\int {\frac {2u\;du}{(3(u^{2}-3)+1)\cdot u}}=-2\int {\frac {du}{3u^{2}-9+1}}=}
=
−
2
∫
d
u
3
u
2
−
8
=
−
2
3
∫
d
u
u
2
−
8
3
.
{\displaystyle =-2\int {\frac {du}{3u^{2}-8}}=-{\frac {2}{3}}\int {\frac {du}{u^{2}-{\frac {8}{3}}}}.}
Šįkart gavome, kad
I
1
{\displaystyle I_{1}}
I
2
{\displaystyle I_{2}}
μ
=
1
,
ν
=
−
1.
{\displaystyle \mu =1,\;\nu =-1.}
https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions )
∫
1
x
2
−
a
2
d
x
=
1
2
a
ln
|
x
−
a
x
+
a
|
+
C
.
{\displaystyle \int {\frac {1}{x^{2}-a^{2}}}\,dx={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C.}
Tęsiame skaičiavimus:
I
1
=
−
2
3
∫
d
u
u
2
−
8
3
=
−
2
3
1
2
8
3
ln
|
u
−
8
3
u
+
8
3
|
=
−
1
3
1
2
2
3
ln
|
u
−
8
3
u
+
8
3
|
=
−
1
6
2
3
ln
|
u
−
8
3
u
+
8
3
|
=
−
1
2
6
ln
|
u
−
8
3
u
+
8
3
|
=
{\displaystyle I_{1}=-{\frac {2}{3}}\int {\frac {du}{u^{2}-{\frac {8}{3}}}}=-{\frac {2}{3}}{\frac {1}{2{\sqrt {\frac {8}{3}}}}}\ln \left|{\frac {u-{\sqrt {\frac {8}{3}}}}{u+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{3}}{\frac {1}{2{\sqrt {\frac {2}{3}}}}}\ln \left|{\frac {u-{\sqrt {\frac {8}{3}}}}{u+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{6{\sqrt {\frac {2}{3}}}}}\ln \left|{\frac {u-{\sqrt {\frac {8}{3}}}}{u+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {u-{\sqrt {\frac {8}{3}}}}{u+{\sqrt {\frac {8}{3}}}}}\right|=}
=
−
1
2
6
ln
|
t
2
+
3
−
8
3
t
2
+
3
+
8
3
|
=
−
1
2
6
ln
|
(
(
1
−
x
)
/
(
1
+
x
)
)
2
+
3
−
8
3
(
(
1
−
x
)
/
(
1
+
x
)
)
2
+
3
+
8
3
|
=
{\displaystyle =-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {t^{2}+3}}-{\sqrt {\frac {8}{3}}}}{{\sqrt {t^{2}+3}}+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {((1-x)/(1+x))^{2}+3}}-{\sqrt {\frac {8}{3}}}}{{\sqrt {((1-x)/(1+x))^{2}+3}}+{\sqrt {\frac {8}{3}}}}}\right|=}
=
−
1
2
6
ln
|
(
1
−
2
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
+
3
−
8
3
(
1
−
2
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
+
3
+
8
3
|
=
−
1
2
6
ln
|
(
1
−
2
x
+
x
2
+
3
+
6
x
+
3
x
2
)
/
(
1
+
2
x
+
x
2
)
−
8
3
(
1
−
2
x
+
x
2
+
3
+
6
x
+
3
x
2
)
/
(
1
+
2
x
+
x
2
)
+
8
3
|
=
{\displaystyle =-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {(1-2x+x^{2})/(1+2x+x^{2})+3}}-{\sqrt {\frac {8}{3}}}}{{\sqrt {(1-2x+x^{2})/(1+2x+x^{2})+3}}+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {(1-2x+x^{2}+3+6x+3x^{2})/(1+2x+x^{2})}}-{\sqrt {\frac {8}{3}}}}{{\sqrt {(1-2x+x^{2}+3+6x+3x^{2})/(1+2x+x^{2})}}+{\sqrt {\frac {8}{3}}}}}\right|=}
=
−
1
2
6
ln
|
(
4
+
4
x
+
4
x
2
)
/
(
1
+
2
x
+
x
2
)
−
8
3
(
4
+
4
x
+
4
x
2
)
/
(
1
+
2
x
+
x
2
)
+
8
3
|
=
−
1
2
6
ln
|
2
(
1
+
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
−
2
2
3
2
(
1
+
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
+
2
2
3
|
=
{\displaystyle =-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {(4+4x+4x^{2})/(1+2x+x^{2})}}-{\sqrt {\frac {8}{3}}}}{{\sqrt {(4+4x+4x^{2})/(1+2x+x^{2})}}+{\sqrt {\frac {8}{3}}}}}\right|=-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {2{\sqrt {(1+x+x^{2})/(1+2x+x^{2})}}-2{\sqrt {\frac {2}{3}}}}{2{\sqrt {(1+x+x^{2})/(1+2x+x^{2})}}+2{\sqrt {\frac {2}{3}}}}}\right|=}
=
−
1
2
6
ln
|
(
1
+
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
−
2
3
(
1
+
x
+
x
2
)
/
(
1
+
2
x
+
x
2
)
+
2
3
|
=
−
1
2
6
ln
|
1
+
x
+
x
2
(
1
+
x
)
2
−
2
3
1
+
x
+
x
2
(
1
+
x
)
2
+
2
3
|
.
{\displaystyle =-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {(1+x+x^{2})/(1+2x+x^{2})}}-{\sqrt {\frac {2}{3}}}}{{\sqrt {(1+x+x^{2})/(1+2x+x^{2})}}+{\sqrt {\frac {2}{3}}}}}\right|=-{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}-{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}+{\sqrt {\frac {2}{3}}}}}\right|.}
Kaip matome, gautas integralas
I
1
{\displaystyle I_{1}}
I
2
{\displaystyle I_{2}}
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1}
Tik dar pastebėsime, kad iš lygybės
ln
a
b
=
−
ln
(
a
b
)
−
1
=
−
ln
b
a
{\displaystyle \ln {\frac {a}{b}}=-\ln({\frac {a}{b}})^{-1}=-\ln {\frac {b}{a}}}
−
1
2
6
ln
|
1
+
x
+
x
2
(
1
+
x
)
2
−
2
3
1
+
x
+
x
2
(
1
+
x
)
2
+
2
3
|
=
1
2
6
ln
|
1
+
x
+
x
2
(
1
+
x
)
2
+
2
3
1
+
x
+
x
2
(
1
+
x
)
2
−
2
3
|
.
{\displaystyle -{\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}-{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}+{\sqrt {\frac {2}{3}}}}}\right|={\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}-{\sqrt {\frac {2}{3}}}}}\right|.}
Integralui
I
2
{\displaystyle I_{2}}
v
=
1
+
3
t
2
,
v
2
=
1
+
3
t
2
,
v
2
3
=
1
3
+
1
t
2
,
v
2
3
−
1
3
=
1
t
2
,
{\displaystyle v={\sqrt {1+{\frac {3}{t^{2}}}}},\;\;v^{2}=1+{\frac {3}{t^{2}}},\;\;{\frac {v^{2}}{3}}={\frac {1}{3}}+{\frac {1}{t^{2}}},\;\;{\frac {v^{2}}{3}}-{\frac {1}{3}}={\frac {1}{t^{2}}},}
d
(
1
t
2
)
=
d
(
v
2
3
−
1
3
)
=
2
3
v
d
v
.
{\displaystyle d{\Big (}{\frac {1}{t^{2}}}{\Big )}=d{\Big (}{\frac {v^{2}}{3}}-{\frac {1}{3}}{\Big )}={\frac {2}{3}}v\;dv.}
Skaičiuojame
I
2
{\displaystyle I_{2}}
I
2
=
−
2
∫
d
t
(
3
t
2
+
1
)
t
2
+
3
=
−
2
∫
d
t
t
2
(
3
+
1
t
2
)
t
1
+
3
t
2
=
−
2
∫
1
t
3
d
t
(
3
+
1
t
2
)
1
+
3
t
2
=
{\displaystyle I_{2}=-2\int {\frac {dt}{(3t^{2}+1){\sqrt {t^{2}+3}}}}=-2\int {\frac {dt}{t^{2}(3+{\frac {1}{t^{2}}})t{\sqrt {1+{\frac {3}{t^{2}}}}}}}=-2\int {\frac {{\frac {1}{t^{3}}}dt}{(3+{\frac {1}{t^{2}}}){\sqrt {1+{\frac {3}{t^{2}}}}}}}=}
=
−
2
−
2
∫
d
(
1
t
2
)
(
3
+
1
t
2
)
1
+
3
t
2
=
∫
d
(
1
t
2
)
(
3
+
1
t
2
)
1
+
3
t
2
=
∫
2
3
v
d
v
(
3
+
v
2
3
−
1
3
)
⋅
v
=
2
3
∫
d
v
3
+
v
2
3
−
1
3
=
{\displaystyle ={\frac {-2}{-2}}\int {\frac {d({\frac {1}{t^{2}}})}{(3+{\frac {1}{t^{2}}}){\sqrt {1+{\frac {3}{t^{2}}}}}}}=\int {\frac {d({\frac {1}{t^{2}}})}{(3+{\frac {1}{t^{2}}}){\sqrt {1+{\frac {3}{t^{2}}}}}}}=\int {\frac {{\frac {2}{3}}v\;dv}{(3+{\frac {v^{2}}{3}}-{\frac {1}{3}})\cdot v}}={\frac {2}{3}}\int {\frac {dv}{3+{\frac {v^{2}}{3}}-{\frac {1}{3}}}}=}
=
2
3
∫
d
v
9
−
1
3
+
v
2
3
=
2
3
∫
d
v
8
+
v
2
3
=
2
∫
d
v
8
+
v
2
=
2
∫
d
v
8
(
1
+
v
2
8
)
=
2
8
∫
d
v
1
+
v
2
8
=
1
4
∫
8
d
(
v
8
)
1
+
v
2
8
=
{\displaystyle ={\frac {2}{3}}\int {\frac {dv}{{\frac {9-1}{3}}+{\frac {v^{2}}{3}}}}={\frac {2}{3}}\int {\frac {dv}{\frac {8+v^{2}}{3}}}=2\int {\frac {dv}{8+v^{2}}}=2\int {\frac {dv}{8(1+{\frac {v^{2}}{8}})}}={\frac {2}{8}}\int {\frac {dv}{1+{\frac {v^{2}}{8}}}}={\frac {1}{4}}\int {\frac {{\sqrt {8}}d{\big (}{\frac {v}{\sqrt {8}}}{\big )}}{1+{\frac {v^{2}}{8}}}}=}
=
8
4
arctan
v
8
+
C
=
2
2
4
arctan
v
8
+
C
=
2
2
arctan
v
8
+
C
=
1
2
arctan
1
+
3
t
2
8
+
C
=
{\displaystyle ={\frac {\sqrt {8}}{4}}\arctan {\frac {v}{\sqrt {8}}}+C={\frac {2{\sqrt {2}}}{4}}\arctan {\frac {v}{\sqrt {8}}}+C={\frac {\sqrt {2}}{2}}\arctan {\frac {v}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3}{t^{2}}}}}{\sqrt {8}}}+C=}
=
1
2
arctan
1
+
3
[
(
1
−
x
)
/
(
1
+
x
)
]
2
8
+
C
=
1
2
arctan
1
+
3
(
1
+
x
)
2
(
1
−
x
)
2
8
+
C
=
1
2
arctan
1
+
3
(
1
+
2
x
+
x
2
)
1
−
2
x
+
x
2
8
+
C
=
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3}{[(1-x)/(1+x)]^{2}}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3(1+x)^{2}}{(1-x)^{2}}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3(1+2x+x^{2})}{1-2x+x^{2}}}}}{\sqrt {8}}}+C=}
=
1
2
arctan
1
−
2
x
+
x
2
+
3
(
1
+
2
x
+
x
2
)
1
−
2
x
+
x
2
8
+
C
=
1
2
arctan
4
+
4
x
+
4
x
2
1
−
2
x
+
x
2
8
+
C
=
1
2
arctan
1
+
x
+
x
2
1
−
2
x
+
x
2
2
+
C
=
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {\frac {1-2x+x^{2}+3(1+2x+x^{2})}{1-2x+x^{2}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {\frac {4+4x+4x^{2}}{1-2x+x^{2}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {\frac {1+x+x^{2}}{1-2x+x^{2}}}}{\sqrt {2}}}+C=}
=
1
2
arctan
1
+
x
+
x
2
(
1
−
x
)
2
2
+
C
=
1
2
arctan
1
+
x
+
x
2
2
(
1
−
x
)
2
+
C
.
{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {\frac {1+x+x^{2}}{(1-x)^{2}}}}{\sqrt {2}}}+C={\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {1+x+x^{2}}{2(1-x)^{2}}}}+C.}
Taigi, dabar gavome, kad integralas
I
2
{\displaystyle I_{2}}
I
1
{\displaystyle I_{1}}
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1}
Galutinai
I
=
∫
d
x
(
x
2
−
x
+
1
)
x
2
+
x
+
1
=
I
1
+
I
2
=
1
2
6
ln
|
1
+
x
+
x
2
(
1
+
x
)
2
+
2
3
1
+
x
+
x
2
(
1
+
x
)
2
−
2
3
|
+
1
2
arctan
1
+
x
+
x
2
2
(
1
−
x
)
2
+
C
.
{\displaystyle I=\int {\frac {dx}{(x^{2}-x+1){\sqrt {x^{2}+x+1}}}}=I_{1}+I_{2}={\frac {1}{2{\sqrt {6}}}}\ln \left|{\frac {{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}+{\sqrt {\frac {2}{3}}}}{{\sqrt {\frac {1+x+x^{2}}{(1+x)^{2}}}}-{\sqrt {\frac {2}{3}}}}}\right|+{\frac {1}{\sqrt {2}}}\arctan {\sqrt {\frac {1+x+x^{2}}{2(1-x)^{2}}}}+C.}
Graži teorija, bet praktikoje ji neveikia kaip aptarta/parodyta aukščiau.
- naudotojas Paraboloid [2024 kovo 2]. 
![{\displaystyle ={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2a\mu \nu +b\mu +b\nu +2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9289d3775ab86af30ed859094bab0ec1fc647734)
![{\displaystyle ={\frac {(a\mu ^{2}+b\mu +c)t^{2}+[2\mu \nu a+b(\mu +\nu )+2c]t+(a\nu ^{2}+b\nu +c)}{(1+t)^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9042cba6daa94f6a97f7a246c54d67188fea07d)
![{\displaystyle x^{2}+px+q={\frac {(\mu ^{2}+p\mu +q)t^{2}+[2\mu \nu +p(\mu +\nu )+2q]t+(\nu ^{2}+p\nu +q)}{(1+t)^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd2089d8aa52927f96a9950776f947e2a04b870a)
![{\displaystyle x^{2}+x+1={\frac {(\mu ^{2}+\mu +1)t^{2}+[2\mu \nu +(\mu +\nu )+2]t+(\nu ^{2}+\nu +1)}{(1+t)^{2}}}\;,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b0ddd873d3b5123eb46819b9631cd90c2b37cb9)
![{\displaystyle x^{2}-x+1={\frac {(\mu ^{2}-\mu +1)t^{2}+[2\mu \nu -(\mu +\nu )+2]t+(\nu ^{2}-\nu +1)}{(1+t)^{2}}}\;.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d57cec0a8425d1944e1e94da34153d14055539d2)
![{\displaystyle ={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3}{[(1-x)/(1+x)]^{2}}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3(1+x)^{2}}{(1-x)^{2}}}}}{\sqrt {8}}}+C={\frac {1}{\sqrt {2}}}\arctan {\frac {\sqrt {1+{\frac {3(1+2x+x^{2})}{1-2x+x^{2}}}}}{\sqrt {8}}}+C=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6791f6fc6300f86c9c18f06b790f371335ee0ab)
