Trečiojo tipo integralo pavyzdyje gavome, kad
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}.}
Apskaičiuosime kam lygus
I
1
,
{\displaystyle I_{1},}
kai 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},}
kai 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}}
ir
I
2
{\displaystyle I_{2}}
integralus, kai x kinta nuo 0 iki 3, gauname:
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}}}}}
reikšme (kai x=0 ir kai x=3) padauginta iš 3. Tada gausime maksimalią ir minimalią įmanomą integralo 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}\;}
arba
0.11886432776254909757535894288364 < 0.35110442407196159407207824841934 < 3.
Vadinasi, integralo I skaičiavime didelių klaidų nėra ir gautas atsakymas (
I
≈
0.35110442407196
{\displaystyle I\approx 0.35110442407196}
) pilnai pretenduoja į teisingo atsakymo statusą.
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}}}}}
nuo 0 iki 3 lygus
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}}}}
vietoje
2
3
,
{\displaystyle {\sqrt {\frac {2}{3}}},}
bet tada gaunama dar mažesnė integralo I reikšmė (integralas
I
2
{\displaystyle I_{2}}
gaunamas neigiamas, va toks:
I
2
{\displaystyle I_{2}}
= -0.03734521977784111855980402563914).
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)}
yra atvirkštinė hiperbolinio tangento funkcija. Matome, kad Wolframalpha integratorius nemoka integruoti iracionaliųjų funkcijų, nes duoda atsakymą su menamuoju vienetu.
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}}}},}
bet tas integralas nereikš ploto po funkcija f(x). Ir taip gali buti su daug integruotų iracionaliųjų funkcijų šituo budu ar taikant Oilerio keitinius.
Kad integravimo keliu gautas atsakymas
I
≈
0.35110442407196
{\displaystyle I\approx 0.35110442407196}
yra neteisingas (integruojant nuo 0 iki 3), galima įsitikinti paėmus f(1) padaugintą iš 1:
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}}}}}
turi didesnes reikšmes (nei f(1)), o mes paėmėme, kad plotas
S
1
{\displaystyle S_{1}}
yra plotas stačiakampio su kraštinėmis 1 ir ~0.577350269. Taigi,
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 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}}}
kreive, kai x kinta nuo 0 iki 0.8:
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}}\;}
į polinomą
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c}
ir pažiūrėsime ar gausime tą pačią išraišką kaip trečiojo tipo integralo teorijoje. Taigi,
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}
įstatę
a
=
1
,
b
=
p
,
c
=
q
{\displaystyle a=1,\;b=p,\;c=q\;}
gausime kam lygus
x
2
+
p
x
+
q
{\displaystyle x^{2}+px+q}
įstačius
x
=
μ
t
+
ν
1
+
t
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;}
keitinį:
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}
išraišką kaip teorijoje. Skaičiavimuose klaidų nėra.
Kadangi
x
=
μ
t
+
ν
1
+
t
,
{\displaystyle x={\frac {\mu t+\nu }{1+t}}\;,}
tai
(
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}}}}.}
Tai 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)}
). Po tokio pakeitimo
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 }
ir
ν
{\displaystyle \nu }
randame iš lygčių sistemos
{
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\;}
(gali būti ir priešingai:
μ
=
−
1
,
ν
=
1
{\displaystyle \mu =-1,\;\nu =1}
). Šįkart mes imsime, kad
μ
=
−
1
,
ν
=
1.
{\displaystyle \mu =-1,\;\nu =1.}
Vadinasi, (7.76) keitinys yra
x
=
μ
t
+
ν
1
+
t
=
−
t
+
1
t
+
1
,
{\displaystyle x={\frac {\mu t+\nu }{1+t}}={\frac {-t+1}{t+1}}\;,}
todėl
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}}
apskaičiuoti naudosime keitinį
u
=
t
2
+
3
,
{\displaystyle u={\sqrt {t^{2}+3}},\;}
o integralui
I
2
{\displaystyle I_{2}}
- keitinį
v
=
1
+
3
t
2
.
{\displaystyle v={\sqrt {1+{\frac {3}{t^{2}}}}}.}
Tada integralui
I
1
{\displaystyle I_{1}}
turime:
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\;}
ir
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}}
integralas yra lygiai toks pats kaip skaičiuotas puslapyje integralas
I
2
{\displaystyle I_{2}}
su
μ
=
1
,
ν
=
−
1.
{\displaystyle \mu =1,\;\nu =-1.}
Iš integralų lentelės ( 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}}
lygus integralui
I
2
{\displaystyle I_{2}}
iš puslapio trečiojo tipo integralo pavyzdžio su
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1}
reikšmėm.
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}}}
turime:
−
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}}
turime:
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}}
integralą:
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}}
lygus integralui
I
1
{\displaystyle I_{1}}
iš puslapio trečiojo tipo integralo pavyzdžio su
μ
=
1
,
ν
=
−
1
{\displaystyle \mu =1,\;\nu =-1}
reikšmėm.
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].