Matematika/Iracionaliųjų funkcijų integravimas

Šis straipsnis yra apie iracionaliųjų funkcijų integravimą.

• $\int {\frac {\sqrt {x}}{x^{\frac {3}{4}}+1}}dx=\int {\frac {u^{2}\cdot 4u^{3}du}{u^{3}+1}}=4[\int u^{2}du-\int {\frac {u^{2}du}{u^{3}+1}}]=4[{\frac {u^{3}}{3}}-{\frac {1}{3}}\int {\frac {d(u^{3}+1)}{u^{3}+1}}]=$

$={\frac {4}{3}}[u^{3}-\ln(u^{3}+1)]+C={\frac {4}{3}}[x^{\frac {3}{4}}-\ln(x^{\frac {3}{4}}+1)]+C,$ kur $x=u^{4};$ $dx=4u^{3}du.$

• $\int {\frac {x+x^{\frac {2}{3}}+x^{\frac {1}{6}}}{x(1+x^{\frac {1}{3}})}}dx=6\int {\frac {(t^{6}+t^{4}+t)t^{5}}{t^{6}(1+t^{2})}}dt=6\int {\frac {t^{5}+t^{3}+1}{t^{2}+1}}dt=6\int t^{3}dt+6\int {\frac {dt}{t^{2}+1}}=$

$={\frac {3}{2}}t^{4}+6\arctan t+C={\frac {3}{2}}x^{\frac {2}{3}}+6\arctan x^{\frac {1}{6}}+C,$ kur $x=t^{6};$ $dx=6t^{5}dt.$

• $\int {\frac {2}{(2-x)^{2}}}({\frac {2-x}{2+x}})^{\frac {1}{3}}dx=-\int {\frac {2(1+t^{3})^{2}t\cdot 12t^{2}}{16t^{6}(1+t^{3})^{2}}}dt=-{\frac {3}{2}}\int {\frac {dt}{t^{3}}}={\frac {3}{4t^{2}}}+C={\frac {3}{4}}({\frac {2+x}{2-x}})^{\frac {2}{3}}+C,$ kur ${\frac {2-x}{2+x}}=t^{3};$ $x={\frac {2-t^{3}}{1+t^{3}}};$ $2-x={\frac {4t^{3}}{1+t^{3}}};$ $-dx={12t^{2}(1+t^{3})-4t^{3}\cdot 3t^{2} \over (1+t^{3})^{2}}dt={12t^{2}+12t^{5}-12t^{5} \over (1+t^{3})^{2}}dt={12t^{2} \over (1+t^{3})^{2}}dt;$ $dx={\frac {-12t^{2}}{(1+t^{3})^{2}}}dt.$

• $\int {\frac {dx}{((x-1)^{3}(x+2)^{5})^{\frac {1}{4}}}}=-\int {\frac {(t^{4}-1)(t^{4}-1)12t^{3}dt}{3\cdot 3t^{4}t(t^{4}-1)^{2}}}=-{\frac {4}{3}}\int {\frac {dt}{t^{2}}}={\frac {4}{3t}}+C={\frac {4}{3}}({\frac {x-1}{x+2}})^{\frac {1}{4}}+C,$ kur $((x-1)^{3}(x+2)^{5})^{\frac {1}{4}}=(x-1)(x+2)({\frac {x+2}{x-1}})^{\frac {1}{4}};$ ${\frac {x+2}{x-1}}=t^{4};$ $x={\frac {t^{4}+2}{t^{4}-1}};$ $x-1={\frac {3}{t^{4}-1}};$ $x+2={\frac {3t^{4}}{t^{4}-1}};$ $dx={\frac {-12t^{3}}{(t^{4}-1)^{2}}}dt.$

• $\int {\frac {1}{(1-x)^{2}}}({\frac {1+x}{1-x}})^{\frac {1}{3}}dx=\int {\frac {(1+u^{3})^{2}u\cdot 6u^{2}}{4(1+u^{3})^{2}}}du={\frac {3}{2}}\cdot {\frac {u^{4}}{4}}+C={\frac {3}{8}}\cdot {\frac {1+x}{1-x}}({\frac {1+x}{1-x}})^{\frac {1}{3}}+C,$ kur ${\frac {1+x}{1-x}}=u^{3};$ $x={\frac {u^{3}-1}{u^{3}+1}};$ $dx={\frac {6u^{2}}{(1+u^{3})^{2}}}du;$ $1-x={\frac {2}{1+u^{3}}}.$

Papildomai

• Iracionaliųjų funkcijų integravimas (papildomai)

Oilerio keitiniai

Funkcijos

$R(x,\;{\sqrt {ax^{2}+bx+c}})\quad (7.66)$

(a, b ir c - realieji skačiai) integralas yra elementarioji funkcija.

I. ${\sqrt {ax^{2}+bx+c}}=u\pm x{\sqrt {a}}$ ; a>0;

$u={\sqrt {ax^{2}+bx+c}}+x{\sqrt {a}}$ . Pakėlus abi dalis lygybės ${\sqrt {ax^{2}+bx+c}}=u-x{\sqrt {a}}$ kvadartu, gauname $bx+c=u^{2}-2{\sqrt {a}}ux,$ taip kad

$x={\frac {u^{2}-c}{2{\sqrt {a}}u+b}},$

${\sqrt {ax^{2}+bx+c}}=u-{\frac {u^{2}-c}{2{\sqrt {a}}u+b}}\cdot {\sqrt {a}}={\frac {u(2{\sqrt {a}}u+b)-({\sqrt {a}}u^{2}-{\sqrt {a}}c)}{2{\sqrt {a}}u+b}}={\frac {{\sqrt {a}}u^{2}+bu+c{\sqrt {a}}}{2{\sqrt {a}}u+b}},$

$dx=({\frac {u^{2}-c}{2{\sqrt {a}}u+b}})'={\frac {2u\cdot (2{\sqrt {a}}u+b)-(u^{2}-c)\cdot 2{\sqrt {a}}}{(2{\sqrt {a}}u+b)^{2}}}du=2{\frac {{\sqrt {a}}u^{2}+bu+c{\sqrt {a}}}{(2{\sqrt {a}}u+b)^{2}}}\;du.$

Tinka, kai kvadrainis trinaris turi menamas šaknis.

II. ${\sqrt {ax^{2}+bx+c}}=xt\pm {\sqrt {c}}$ ; c>0. Tada

$ax^{2}+bx+c=x^{2}t^{2}+2xt{\sqrt {c}}+c.$

(Mes apsisprendėme prieš šaknį pasirinkti pliuso ženklą.) Iš čia x yra kaip racionali funkcija nuo t:

$ax^{2}-x^{2}t^{2}+bx-2xt{\sqrt {c}}=0,$

$(a-t^{2})x^{2}+(b-2t{\sqrt {c}})x=0,$

$x((a-t^{2})x+b-2t{\sqrt {c}})=0;$ iš čia arba x=0 arba $(a-t^{2})x+b-2t{\sqrt {c}}=0;$ Mums reikia išreikšti x per t, todėl x=0 netinka; sprendžiame toliau:

$(a-t^{2})x+b-2t{\sqrt {c}}=0;$

$(a-t^{2})x=2{\sqrt {c}}t-b;$

$x={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}.$

Turime

${\sqrt {ax^{2}+bx+c}}=xt+{\sqrt {c}}={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}\cdot t+{\sqrt {c}}={\frac {2{\sqrt {c}}t^{2}-bt}{a-t^{2}}}+{\sqrt {c}};$

$\left({\frac {2{\sqrt {c}}t-b}{a-t^{2}}}\right)'={\frac {(2{\sqrt {c}}t-b)'(a-t^{2})-(2{\sqrt {c}}t-b)(a-t^{2})'}{(a-t^{2})^{2}}}={\frac {2{\sqrt {c}}(a-t^{2})-(2{\sqrt {c}}t-b)(-2t)}{(a-t^{2})^{2}}}={\frac {2a{\sqrt {c}}-2{\sqrt {c}}t^{2}-(-4{\sqrt {c}}t^{2}+2bt)}{(a-t^{2})^{2}}}=$

$={\frac {2a{\sqrt {c}}-2{\sqrt {c}}t^{2}+4{\sqrt {c}}t^{2}-2bt}{(a-t^{2})^{2}}}={\frac {(4{\sqrt {c}}-2{\sqrt {c}})t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}.$

$dx={\frac {2{\sqrt {c}}t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}dt.$

III. ${\sqrt {a(x-x_{1})(x-x_{2})}}=u(x-x_{1}),$ kur $x_{1}$ yra bet kuri realioji trinario $ax^{2}+bx+c$ šaknis. Taikoma tik kai yra du lygties ${\sqrt {ax^{2}+bx+c}}={\sqrt {a(x-x_{1})(x-x_{2})}}$ sprendiniai.

Tegu $\alpha$ ir $\beta$ - realiosios šaknys trinario $ax^{2}+bx+c.$ Tada

${\sqrt {ax^{2}+bx+c}}=(x-\alpha )t.$

Kadangi $ax^{2}+bx+c=a(x-\alpha )(x-\beta ),$ tai

${\sqrt {a(x-\alpha )(x-\beta )}}=(x-\alpha )t,$

$a(x-\alpha )(x-\beta )=(x-\alpha )^{2}t^{2},$

$a(x-\beta )=(x-\alpha )t^{2}.$

Iš čia randame x kaip racionalią funkciją nuo t:

$ax-a\beta =xt^{2}-\alpha t^{2},$

$ax-xt^{2}=a\beta -\alpha t^{2},$

$(a-t^{2})x=a\beta -\alpha t^{2},$

$x={\frac {a\beta -\alpha t^{2}}{a-t^{2}}}.$

Trečias Oilerio keitinys tinka kai a>0 ir kai a<0. Butina tik, kad turėtų daugianaris $ax^{2}+bx+c$ dvi realiasias šaknis.

$({\frac {a\beta -\alpha t^{2}}{a-t^{2}}})'={\frac {-2\alpha t(a-t^{2})-(a\beta -\alpha t^{2})(-2t)}{(a-t^{2})^{2}}}={\frac {-2\alpha at+2\alpha t^{3}-(-2a\beta t+2\alpha t^{3})}{(a-t^{2})^{2}}}=$

$={\frac {-2\alpha at+2\alpha t^{3}+2a\beta t-2\alpha t^{3}}{(a-t^{2})^{2}}}={\frac {-2\alpha at+2a\beta t}{(a-t^{2})^{2}}}={\frac {(2a\beta -2a\alpha )t}{(a-t^{2})^{2}}}.$

$dx={\frac {2a(\beta -\alpha )t}{(a-t^{2})^{2}}}dt.$

Pavyzdžiai

Pirmojo Oilerio keitinio pavyzdžiai

• $\int {\frac {dx}{\sqrt {5x^{2}+x+1}}};$

${\sqrt {5x^{2}+x+1}}=u+x{\sqrt {5}}.$ Pakėlę šios lygybės abi puses kvadratu, gauname: $5x^{2}+x+1=u^{2}+2ux{\sqrt {5}}+5x^{2};\;x+1=u^{2}+2{\sqrt {5}}ux;\;x(1-2u{\sqrt {5}})=u^{2}-1;\;x={\frac {u^{2}-1}{1-2{\sqrt {5}}u}};$

$dx=({\frac {u^{2}-1}{1-2{\sqrt {5}}u}})'={\frac {(u^{2}-1)'(1-2{\sqrt {5}}u)-(u^{2}-1)(1-2{\sqrt {5}}u)'}{(1-2{\sqrt {5}}u)^{2}}}=$

$={\frac {2u(1-2{\sqrt {5}}u)-(u^{2}-1)(-2{\sqrt {5}})}{1-4{\sqrt {5}}u+4\cdot 5\cdot u^{2}}}du={\frac {2u-4{\sqrt {5}}u^{2}+2{\sqrt {5}}(u^{2}-1)}{1-4{\sqrt {5}}u+20u^{2}}}du=$

$={\frac {2u-4{\sqrt {5}}u^{2}+2{\sqrt {5}}\cdot u^{2}-2{\sqrt {5}}}{1-4{\sqrt {5}}u+20u^{2}}}du={\frac {2u-2{\sqrt {5}}u^{2}-2{\sqrt {5}}}{1-4{\sqrt {5}}u+20u^{2}}}du={\frac {2(u-{\sqrt {5}}u^{2}-{\sqrt {5}})}{(1-2{\sqrt {5}}\cdot u)^{2}}}du.$

${\sqrt {5x^{2}+x+1}}=u+x{\sqrt {5}}=u+{\frac {u^{2}-1}{1-2{\sqrt {5}}u}}\cdot {\sqrt {5}}=u+{\frac {{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}={\frac {u(1-2{\sqrt {5}}u)+{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}=$

$={\frac {u-2{\sqrt {5}}u^{2}+{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}={\frac {u-{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}.$

$\int {\frac {dx}{\sqrt {5x^{2}+x+1}}}=\int {\frac {{\frac {2(u-{\sqrt {5}}u^{2}-{\sqrt {5}})}{(1-2{\sqrt {5}}\cdot u)^{2}}}dt}{\frac {u-{\sqrt {5}}u^{2}-{\sqrt {5}}}{1-2{\sqrt {5}}u}}}=2\int {\frac {du}{1-2{\sqrt {5}}u}}=-{\frac {2}{2{\sqrt {5}}}}\int {\frac {d(1-u2{\sqrt {5}})}{1-2{\sqrt {5}}u}}=$

$=-{\frac {1}{\sqrt {5}}}\ln |1-2{\sqrt {5}}u|+C=-{\frac {1}{\sqrt {5}}}\ln |1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x)|+C=$ $=-{\frac {1}{\sqrt {5}}}\ln |1-{\sqrt {100x^{2}+20x+20}}+10x|+C.$

• $\int {\frac {dx}{\sqrt {x^{2}+3}}}.$ Taikome I Oilerio keitinį ${\sqrt {x^{2}+3}}=-x+t=t-x.$

$x^{2}+3=(-x+t)^{2}=x^{2}-2tx+t^{2}$ ; $3=t^{2}-2tx$ ; $2tx=t^{2}-3$ ; $x={\frac {t^{2}-3}{2t}}.$ $dx=({\frac {t^{2}-3}{2t}})'={\frac {(t^{2}-3)'\cdot 2t-(t^{2}-3)\cdot (2t)'}{(2t)^{2}}}={\frac {2t\cdot 2t-(t^{2}-3)\cdot 2}{4t^{2}}}dt={\frac {2t^{2}-(t^{2}-3)}{2t^{2}}}dt={\frac {t^{2}+3}{2t^{2}}}dt.$

${\sqrt {x^{2}+3}}=-x+t=-{\frac {t^{2}-3}{2t}}+t={\frac {-(t^{2}-3)+2t^{2}}{2t}}={\frac {t^{2}+3}{2t}}.$

$\int {\frac {dx}{\sqrt {x^{2}+3}}}=\int {\frac {{\frac {t^{2}+3}{2t^{2}}}dt}{\frac {t^{2}+3}{2t}}}=\int {\frac {dt}{t}}=\ln |t|+C=\ln |x+{\sqrt {x^{2}+3}}|+C.$

• Apskaičiuoti $\int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}.$

Sprendimas. Kadangi trinaris $x^{2}+x+1$ turi kompleksines šaknis, padarysime keitinį ${\sqrt {x^{2}+x+1}}=t-x.$ Pakėlę abi lygybės puses kvadratu, gauname

$x^{2}+x+1=t^{2}-2tx+x^{2}$ arba $x+1=t^{2}-2tx$; iš čia $x={\frac {t^{2}-1}{1+2t}}$ , $dx=({\frac {t^{2}-1}{1+2t}})'={\frac {2t(1+2t)-(t^{2}-1)\cdot 2}{(1+2t)^{2}}}={\frac {2t+4t^{2}-2t^{2}+2}{(1+2t)^{2}}}={\frac {2t+2t^{2}+2}{(1+2t)^{2}}}.$

Tada

$\int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}=\int {\frac {2t+2t^{2}+2}{t(1+2t)^{2}}}dt.$

Toliau, turime

${\frac {2t^{2}+2t+2}{t(1+2t)^{2}}}={\frac {A}{t}}+{\frac {B}{1+2t}}+{\frac {D}{(1+2t)^{2}}}.$

Padauginę abi dalis lygybės su $t(1+2t)^{2},$ gauname

$2t^{2}+2t+2=A(1+2t)^{2}+Bt(1+2t)+Dt=A(1+4t+4t^{2})+Bt+2Bt^{2}+Dt=(4A+2B)t^{2}+(4A+B+D)t+A.$

Prilyginę koeficientus prie vienodų laipsmių t, gauname sistemą lygčių pirmojo laipsnio atžvilgiu A, B, D:

{4A+2B=2,

{4A+B+D=2,

{A=2,

Iš čia A=2, B=-3, D=-3. Todėl,

${\frac {2t^{2}+2t+2}{t(1+2t)^{2}}}={\frac {2}{t}}-{\frac {3}{1+2t}}-{\frac {3}{(1+2t)^{2}}},$

ir galutinai

$\int {\frac {dx}{x+{\sqrt {x^{2}+x+1}}}}=\int [{\frac {2}{t}}-{\frac {3}{1+2t}}-{\frac {3}{(1+2t)^{2}}}]dt=2\int {\frac {dt}{t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{1+2t}}-{\frac {3}{2}}\int {\frac {d(1+2t)}{(1+2t)^{2}}}=$

$=2\ln |t|-{\frac {3}{2}}\ln |1+2t|+{\frac {3}{2(1+2t)}}+C=$

$=2\ln |x+{\sqrt {x^{2}+x+1}}|-{\frac {3}{2}}\ln |1+2x+2{\sqrt {x^{2}+x+1}}|+{\frac {3}{2+4x+4{\sqrt {x^{2}+x+1}}}}+C.$

Antrojo Oilerio keitinio pavyzdžiai

• $\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}};$

${\sqrt {1-6x-9x^{2}}}=ux+1.$ Pakelę abi puses kvadratu, gauname $1-6x-9x^{2}=u^{2}x^{2}+2ux+1;$ $-6x-9x^{2}=u^{2}x^{2}+2ux;$ $-6-9x=u^{2}x+2u.$ Imdami apiejų lygybės pusių diferencialus, randame: $-9dx=2uxdu+2du;$ $dx={\frac {-2(ux+1)}{9+u^{2}}}du.$ $\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=-2\int {\frac {(ux+1)du}{(9+u^{2})(ux+1)}}=-2\int {\frac {du}{9+u^{2}}}=-{\frac {2}{3}}\arctan {\frac {u}{3}}+C=$ $=-{\frac {2}{3}}\arctan {\frac {{\sqrt {1-6x-9x^{2}}}-1}{3x}}+C.$

Sprendimas normaliai. Čia a=-9, b=-6, c=1. Tada

$x={\frac {2{\sqrt {c}}t-b}{a-t^{2}}}={\frac {2{\sqrt {1}}t-(-6)}{-9-t^{2}}}={\frac {2t+6}{-9-t^{2}}}.$

${\sqrt {1-6x-9x^{2}}}=xt+1={\frac {2t+6}{-9-t^{2}}}t+1={\frac {2t^{2}+6t}{-9-t^{2}}}+1={\frac {2t^{2}+6t-9-t^{2}}{-9-t^{2}}}={\frac {t^{2}+6t-9}{-9-t^{2}}}.$

$dx={\frac {2{\sqrt {c}}t^{2}-2bt+2a{\sqrt {c}}}{(a-t^{2})^{2}}}dt={\frac {2{\sqrt {1}}t^{2}-2(-6)t+2(-9){\sqrt {1}}}{(-9-t^{2})^{2}}}dt={\frac {2t^{2}+12t-18}{(-9-t^{2})^{2}}}dt={\frac {2(t^{2}+6t-9)}{(-9-t^{2})^{2}}}dt.$

$\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=\int {\frac {{\frac {2(t^{2}+6t-9)}{(-9-t^{2})^{2}}}dt}{\frac {t^{2}+6t-9}{-9-t^{2}}}}=\int {\frac {{\frac {2}{-9-t^{2}}}dt}{1}}=\int {\frac {2dt}{-9-t^{2}}}=-2\int {\frac {dt}{9+t^{2}}}=-{\frac {2}{3}}\arctan {\frac {t}{3}}+C=$

$=-{\frac {2}{3}}\arctan {\frac {{\sqrt {1-6x-9x^{2}}}-1}{3x}}+C.$

• Apskaičiuoti $\int {\frac {dx}{(1+x){\sqrt {1+x-x^{2}}}}}.$

Sprendimas. Čia trinaris $1+x-x^{2}$ turi kompleksines šaknis ir a<0, c>0, pasinaudojame keitiniu ${\sqrt {1+x-x^{2}}}=tx-1.$ Pakėlę abi lygties puses kvadratu, gauname

$1+x-x^{2}=t^{2}x^{2}-2tx+1$, $x-x^{2}=t^{2}x^{2}-2tx$, $1-x=t^{2}x-2t$, $1+2t=t^{2}x+x$, $x(t^{2}+1)=1+2t$, $x={\frac {1+2t}{t^{2}+1}}$. $dx=({\frac {1+2t}{t^{2}+1}})'={\frac {2(t^{2}+1)-(1+2t)\cdot 2t}{(t^{2}+1)^{2}}}dt={\frac {2t^{2}+2-2t-4t^{2}}{(t^{2}+1)^{2}}}dt={\frac {2-2t-2t^{2}}{(t^{2}+1)^{2}}}dt={\frac {2(1-t-t^{2})}{(t^{2}+1)^{2}}}dt.$ ${\sqrt {1+x-x^{2}}}=t\cdot {\frac {1+2t}{t^{2}+1}}-1={\frac {t+2t^{2}-t^{2}-1}{t^{2}+1}}={\frac {t+t^{2}-1}{t^{2}+1}}.$ ${\sqrt {1+x-x^{2}}}=tx-1;\;{\sqrt {1+x-x^{2}}}+1=tx;\;t={\frac {{\sqrt {1+x-x^{2}}}+1}{x}}.$

Tokiu budu,

$\int {\frac {dx}{(1+x){\sqrt {1+x-x^{2}}}}}=\int {\frac {{\frac {2(1-t-t^{2})}{(t^{2}+1)^{2}}}dt}{(1+{\frac {1+2t}{t^{2}+1}}){\frac {t+t^{2}-1}{t^{2}+1}}}}=\int {\frac {{\frac {-2(t^{2}+t-1)}{(t^{2}+1)^{2}}}dt}{{\frac {t^{2}+1+1+2t}{t^{2}+1}}\cdot {\frac {t^{2}+t-1}{t^{2}+1}}}}=\int {\frac {-2dt}{1+t^{2}+2t+1}}=$ $=\int {\frac {-2d(t+1)}{1+(t+1)^{2}}}=-2\arctan(t+1)+C=-2\arctan({\frac {{\sqrt {1+x-x^{2}}}+1}{x}}+1)+C=$ $=-2\arctan {\frac {{\sqrt {1+x-x^{2}}}+1+x}{x}}+C.$

• Reikia apskaičiuoti integralą

$\int {\frac {(1-{\sqrt {1+x+x^{2}}})^{2}}{x^{2}{\sqrt {1+x+x^{2}}}}}dx.$

Sprendimas. Taikome II Oilerio keitinį ${\sqrt {1+x+x^{2}}}=xt+1;$ tada

$1+x+x^{2}=x^{2}t^{2}+2xt+1;$ $x+x^{2}=x^{2}t^{2}+2xt;$ $1+x=xt^{2}+2t;$ $x-xt^{2}=2t-1;$ $x(1-t^{2})=2t-1;$ $x={\frac {2t-1}{1-t^{2}}}.$

$dx={\frac {2(1-t^{2})-(2t-1)\cdot (-2t)}{(1-t^{2})^{2}}}dt={\frac {2-2t^{2}+2t(2t-1)}{(1-t^{2})^{2}}}dt={\frac {2-2t^{2}+4t^{2}-2t}{(1-t^{2})^{2}}}dt={\frac {2t^{2}-2t+2}{(1-t^{2})^{2}}}dt.$

${\sqrt {1+x+x^{2}}}={\frac {2t-1}{1-t^{2}}}\cdot t+1={\frac {2t^{2}-t+(1-t^{2})}{1-t^{2}}}={\frac {t^{2}-t+1}{1-t^{2}}}.$

$1-{\sqrt {1+x+x^{2}}}=1-(xt+1)=-xt=-{\frac {2t-1}{1-t^{2}}}\cdot t={\frac {-2t^{2}+t}{1-t^{2}}}.$

$t={\frac {{\sqrt {1+x+x^{2}}}-1}{x}}.$

Įstate gautas išraiškas į pradinį integralą, randame:

$\int {\frac {(1-{\sqrt {1+x+x^{2}}})^{2}}{x^{2}{\sqrt {1+x+x^{2}}}}}dx=\int {\frac {(-2t^{2}+t)^{2}(1-t^{2})^{2}(1-t^{2})(2t^{2}-2t+2)}{(1-t^{2})^{2}(2t-1)^{2}(t^{2}-t+1)(1-t^{2})^{2}}}dt=2\int {\frac {(-2t^{2}+t)^{2}}{(2t-1)^{2}(1-t^{2})}}dt=2\int {\frac {t^{2}}{1-t^{2}}}dt=$ $=2\int ({\frac {1}{1-t^{2}}}-1)dt=2({\frac {1}{2}}\ln |{\frac {1+t}{1-t}}|-t)+C=$ $=\ln |{\frac {1+{\frac {{\sqrt {1+x+x^{2}}}-1}{x}}}{1-{\frac {{\sqrt {1+x+x^{2}}}-1}{x}}}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=\ln |{\frac {\frac {x+{\sqrt {1+x+x^{2}}}-1}{x}}{\frac {x-{\sqrt {1+x+x^{2}}}+1}{x}}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=$ $=\ln |{\frac {x+{\sqrt {1+x+x^{2}}}-1}{x-{\sqrt {1+x+x^{2}}}+1}}|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C=$ $=\ln |2x+2{\sqrt {1+x+x^{2}}}+1|-{\frac {2({\sqrt {1+x+x^{2}}}-1)}{x}}+C.$

Trečiojo Oilerio keitinio pavyzdžiai

• $\int {\frac {dx}{x{\sqrt {-x^{2}+5x-6}}}}.$ Pastebėję, kad pošaknio trinario šaknys yra 2 ir 3, taikome keitinį ${\sqrt {(x-2)(3-x)}}=u(x-2).$ Pakėlę šios lygybės abi puses kvadratu ir suprastinę iš $(x-2),$ gauname:

$(x-2)(3-x)=u^{2}(x-2)^{2};$ $3-x=u^{2}(x-2);$ $-xu^{2}-x=-2u^{2}-3;$ $x(-u^{2}-1)=-2u^{2}-3;$ $x={\frac {2u^{2}+3}{u^{2}+1}};\;u={\sqrt {\frac {3-x}{x-2}}}$;

$dx={\frac {4u(u^{2}+1)-2u(2u^{2}+3)}{(u^{2}+1)^{2}}}du={\frac {4u^{3}+4u-4u^{3}-6u}{(u^{2}+1)^{2}}}du={\frac {-2u\;du}{(u^{2}+1)^{2}}}.$

${\sqrt {(x-2)(3-x)}}=u(x-2)=u({\frac {2u^{2}+3}{u^{2}+1}}-2)={\frac {2u^{3}+3u-2u(u^{2}+1)}{u^{2}+1}}={\frac {2u^{3}+3u-2u^{3}-2u}{u^{2}+1}}={\frac {u}{u^{2}+1}}.$

$\int {\frac {dx}{x{\sqrt {-x^{2}+5x-6}}}}=\int {\frac {\frac {-2u\;du}{(u^{2}+1)^{2}}}{{\frac {2u^{2}+3}{u^{2}+1}}\cdot {\frac {u}{u^{2}+1}}}}=-2\int {\frac {du}{3+2u^{2}}}=-\int {\frac {du}{{\frac {3}{2}}+u^{2}}}=$

$=-{\frac {1}{\sqrt {\frac {3}{2}}}}\arctan({\frac {u}{\sqrt {\frac {3}{2}}}})+C=-{\sqrt {\frac {2}{3}}}\arctan({\sqrt {\frac {2}{3}}}u)+C=-{\sqrt {\frac {2}{3}}}\arctan({\sqrt {\frac {2}{3}}}\cdot {\sqrt {\frac {3-x}{x-2}}})+C.$

• $\int {\frac {x\;dx}{\sqrt {(7x-10-x^{2})^{3}}}},$ kur $a<0,$ $c<0,$ todėl taikome III Oilerio keitinį. Lygties $7x-10-x^{2}$ sprendiniai yra $x_{1}=2$, $x_{2}=5$; $a=-1.$

${\sqrt {7x-10-x^{2}}}={\sqrt {-1(x-2)(x-5)}}={\sqrt {(x-2)(5-x)}}=(x-2)t.$ $5-x=(x-2)t^{2};$ $-xt^{2}-x=-2t^{2}-5;$ $x={\frac {5+2t^{2}}{1+t^{2}}};$ $dx=({\frac {5+2t^{2}}{1+t^{2}}})'={\frac {4t(1+t^{2})-(5+2t^{2})2t}{(1+t^{2})^{2}}}dt={\frac {4t+4t^{3}-10t-4t^{3}}{(1+t^{2})^{2}}}dt={\frac {-6t}{(1+t^{2})^{2}}}dt.$ ${\sqrt {7x-10-x^{2}}}=(x-2)t=({\frac {5+2t^{2}}{1+t^{2}}}-2)t={\frac {5+2t^{2}-2-2t^{2}}{1+t^{2}}}t={\frac {3t}{1+t^{2}}}.$ $t={\frac {\sqrt {(x-2)(5-x)}}{(x-2)}}={\sqrt {\frac {(5-x)}{(x-2)}}}.$ $\int {\frac {x\;dx}{\sqrt {(7x-10-x^{2})^{3}}}}=\int {\frac {{\frac {5+2t^{2}}{1+t^{2}}}{\frac {-6t}{(1+t^{2})^{2}}}dt}{({\frac {3t}{1+t^{2}}})^{3}}}=\int {\frac {\frac {-6t(5+2t^{2})}{(1+t^{2})^{3}}}{({\frac {3t}{1+t^{2}}})^{3}}}dt=\int {\frac {-6t(5+2t^{2})}{27t^{3}}}dt=\int {\frac {-6(5+2t^{2})}{27t^{2}}}dt=$ $=\int (-{\frac {30}{27t^{2}}}-{\frac {12t^{2}}{27t^{2}}})dt={\frac {1}{27}}\int (-{\frac {30}{t^{2}}}-12)dt={\frac {1}{27}}({\frac {30}{t}}-12t)+C={\frac {1}{27}}({\frac {30}{\sqrt {\frac {(5-x)}{(x-2)}}}}-12{\sqrt {\frac {(5-x)}{(x-2)}}})+C=$ $={\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {(5-x)}{(x-2)}}}+C.$

• Reikia apskaičiuoti integralą

$\int {\frac {dx}{\sqrt {x^{2}+3x-4}}}.$

Sprendimas. Kadangi $x^{2}+3x-4=(x+4)(x-1),$ tai:

${\sqrt {(x+4)(x-1)}}=(x+4)t;$ tada

$(x+4)(x-1)=(x+4)^{2}t^{2}$, $x-1=(x+4)t^{2},$ $x-1=xt^{2}+4t^{2},$ $x-xt^{2}=4t^{2}+1,$ $x={\frac {1+4t^{2}}{1-t^{2}}}.$

$dx={\frac {(1+4t^{2})'\cdot (1-t^{2})-(1+4t^{2})\cdot (1-t^{2})'}{(1-t^{2})^{2}}}={\frac {8t\cdot (1-t^{2})-(1+4t^{2})\cdot (-2t)}{(1-t^{2})^{2}}}dt=$

$={\frac {8t-8t^{3}+2t+8t^{3}}{(1-t^{2})^{2}}}dt={\frac {10t}{(1-t^{2})^{2}}}dt.$

${\sqrt {(x+4)(x-1)}}=(x+4)t=({\frac {1+4t^{2}}{1-t^{2}}}+4)t={\frac {1+4t^{2}+4-4t^{2}}{1-t^{2}}}\cdot t={\frac {5t}{1-t^{2}}}.$

$t={\frac {\sqrt {(x+4)(x-1)}}{(x+4)}}={\sqrt {\frac {(x+4)(x-1)}{(x+4)^{2}}}}={\sqrt {\frac {x-1}{x+4}}}.$

Grįžtant prie pradinio integralo, gauname:

$\int {\frac {dx}{\sqrt {x^{2}+3x-4}}}=\int {\frac {10t(1-t^{2})}{(1-t^{2})^{2}5t}}dt=\int {\frac {2}{1-t^{2}}}dt=\ln |{\frac {1+t}{1-t}}|+C=\ln |{\frac {1+{\sqrt {\frac {x-1}{x+4}}}}{1-{\sqrt {\frac {x-1}{x+4}}}}}|+C=$ $=\ln |{\frac {{\sqrt {x+4}}+{\sqrt {x-1}}}{{\sqrt {x+4}}-{\sqrt {x-1}}}}|+C.$

Diferencialinių binomų integravimas

Integralas $\int x^{m}(a+bx^{n})^{p}dx,$ kur m, n, p - racionalieji skaičiai, vadinamas integralu su binominiu diferencialu. Šį integralą elementariosiomis funkcijomis įmanoma išreikšti tik trimis atvejais:

I. p - sveikasis skaičius. Jei $p>0,$ tai pointegralinis binomas skleidžiamas pagal Niutono binomo formulę. Jei $p<0,$ tai keičiame $x=t^{k},$ kur k - bendras trupmenų m ir n vardiklis. Pavyzdžiui, trupmenų ${\tfrac {1}{4}}$ ir ${\tfrac {2}{3}}$ bendras vardiklis yra 3 $\cdot$ 4 = 12.

II. ${\frac {m+1}{n}}$ - sveikasis skaičius. Keičiame $a+bx^{n}=t^{\alpha },$ kur $\alpha$ - trupmenos p vardiklis.

III. ${\frac {m+1}{n}}+p$ - sveikasis skaičius. Keičiame $a+bx^{n}=t^{\alpha }x^{n},$ kur $\alpha$ - trupmenos p vardiklis.

Pavyzdžiai

• $\int x^{\frac {1}{3}}(2+{\sqrt {x}})^{2}dx=\int x^{\frac {1}{3}}(4+4{\sqrt {x}}+x)dx=\int (4x^{\frac {1}{3}}+4x^{\frac {5}{6}}+x^{\frac {4}{3}})dx=3x^{\frac {4}{3}}+{\frac {24}{11}}x^{\frac {11}{6}}+{\frac {3}{7}}x^{\frac {7}{3}}+C,$

kur $p=2$ - sveikasis skaičius. Turime I atvejį.

• $\int {\frac {\sqrt {1+x^{\frac {1}{3}}}}{x^{\frac {2}{3}}}}dx=\int x^{-{\frac {2}{3}}}(1+x^{\frac {1}{3}})^{\frac {1}{2}}dx=\int {\sqrt {t^{2}}}6t\;dt=6\int t^{2}\;dt=2t^{3}+C=2(1+x^{\frac {1}{3}})^{\frac {3}{2}}+C,$

kur $m=-{\frac {2}{3}};\;n={\frac {1}{3}};\;p={\frac {1}{2}};\;{\frac {m+1}{n}}=1.$ Turime II atvejį. $1+x^{\frac {1}{3}}=t^{2};\;{\frac {1}{3}}x^{-{\frac {2}{3}}}dx=2t\;dt.$

• $\int x^{-11}(1+x^{4})^{-{\frac {1}{2}}}dx=-{\frac {1}{2}}\int (t^{2}-1)^{\frac {11}{4}}({\frac {t^{2}}{t^{2}-1}})^{-{\frac {1}{2}}}{\frac {t\;dt}{(t^{2}-1)^{\frac {5}{4}}}}=-{\frac {1}{2}}\int (t^{2}-1)^{{\frac {11}{4}}+{\frac {1}{2}}-{\frac {5}{4}}}{\frac {t}{t^{\frac {2}{2}}}}dt=$

$=-{\frac {1}{2}}\int (t^{2}-1)^{2}dt=-{\frac {t^{5}}{10}}+{\frac {t^{3}}{3}}-{\frac {t}{2}}+C=-{\frac {(1+x^{4})^{\frac {5}{2}}}{10x^{2}}}+{\frac {(1+x^{4})^{\frac {3}{2}}}{3x^{2}}}-{\frac {(1+x^{4})^{\frac {1}{2}}}{2x^{2}}}+C,$ kur $m=-11;$ $n=4;$ $p=-{\frac {1}{2}};\;{\frac {m+1}{n}}+p=-3.$ Turime III atvejį. $1+x^{4}=x^{4}t^{2};$ $x={\frac {1}{(t^{2}-1)^{\frac {1}{4}}}};\;dx=-{\frac {t\;dt}{2(t^{2}-1)^{\frac {5}{4}}}};\;t={\frac {\sqrt {1+x^{4}}}{x^{2}}}.$

• $\int {\frac {x^{3}\;dx}{\sqrt {1+2x^{2}}}}=\int x^{3}(1+2x^{2})^{-{\frac {1}{2}}}dx={\frac {1}{2}}\int ({\frac {u^{2}-1}{2}})^{\frac {3}{2}}u^{-1}({\frac {u^{2}-1}{2}})^{-{\frac {1}{2}}}u\;du={\frac {1}{2}}\int {\frac {u^{2}-1}{2}}du=$

$={\frac {1}{4}}({\frac {u^{3}}{3}}-u)+C={\frac {(1+2x^{2})^{\frac {3}{2}}}{12}}-{\frac {\sqrt {1+2x^{2}}}{4}}+C,$ kur ${\frac {m+1}{n}}={\frac {3+1}{2}}=2;$ $1+2x^{2}=u^{2};$ $x=({\frac {u^{2}-1}{2}})^{\frac {1}{2}};\;dx={\frac {1}{2}}({\frac {u^{2}-1}{2}})^{-{\frac {1}{2}}}u\;du.$

• $\int x^{-{\frac {1}{3}}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3\int t^{-1}\cdot t\;dt=-3\int dt=-3t+C=-3{\sqrt {1-x^{\frac {2}{3}}}}+C,$

kur $m=-{\frac {1}{3}};$ $n={\frac {2}{3}};\;p=-{\frac {1}{2}};\;{\frac {m+1}{n}}=1;\;1-x^{\frac {2}{3}}=t^{2};\;-{\frac {2}{3}}x^{-{\frac {1}{3}}}dx=2t\;dt;\;dx=-3x^{\frac {1}{3}}tdt;\;x^{\frac {2}{3}}=1-t^{2}.$

• $\int x^{\frac {1}{3}}(1-x^{\frac {2}{3}})^{-{\frac {1}{2}}}dx=-3\int x^{\frac {1}{3}}\cdot (t^{2})^{-{\frac {1}{2}}}\cdot x^{\frac {1}{3}}\cdot t\;dt=-3\int x^{\frac {2}{3}}\cdot t^{-1}\cdot t\;dt=$

$=-3\int (1-t^{2}){\frac {t}{t}}\;dt=-3\int (1-t^{2})\;dt=-3(t-{\frac {t^{3}}{3}})+C=-3({\sqrt {1-x^{\frac {2}{3}}}}-{\frac {\sqrt {(1-x^{\frac {2}{3}})^{3}}}{3}})+C=$ $=-3{\frac {3(1-x^{\frac {2}{3}})^{\frac {1}{2}}-(1-x^{\frac {2}{3}})^{\frac {3}{2}}}{3}}+C=-3(1-x^{\frac {2}{3}})^{\frac {1}{2}}+(1-x^{\frac {2}{3}})^{\frac {3}{2}}+C=$

$=-3{\sqrt {1-x^{\frac {2}{3}}}}+{\sqrt {1-3x^{\frac {2}{3}}+3x^{\frac {4}{3}}-x^{2}}}+C,$

kur $m={\frac {1}{3}};$ $n={\frac {2}{3}};\;p=-{\frac {1}{2}};\;{\frac {m+1}{n}}=2;\;1-x^{\frac {2}{3}}=t^{2};\;-{\frac {2}{3}}x^{-{\frac {1}{3}}}dx=2t\;dt;\;dx=-3x^{\frac {1}{3}}tdt;\;x^{\frac {2}{3}}=1-t^{2}.$

• $\int {\frac {dx}{\sqrt {x^{2}+3}}}.$ Matome, kad tinka trečias atvejis, nes ${\frac {m+1}{n}}+p={\frac {0+1}{2}}-{\frac {1}{2}}=0$. Čia m=0, n=2, $p=-{\frac {1}{2}}$. Keičiame $a+bx^{n}=t^{\alpha }x^{n},$ kur $\alpha$ - trupmenos p vardiklis. Taigi $a+bx^{n}=t^{\alpha }x^{n}$, čia a=3, b=1; $x^{2}+3=t^{2}x^{2}$; $3=t^{2}x^{2}-x^{2}$; $x^{2}={\frac {3}{t^{2}-1}}=3(t^{2}-1)^{-1}$; $x={\sqrt {3}}\cdot (t^{2}-1)^{-{\frac {1}{2}}}$.

$dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}\cdot 2t\;dt=-{\frac {{\sqrt {3}}t}{\sqrt {(t^{2}-1)^{3}}}}dt.$

$\int {\frac {dx}{\sqrt {x^{2}+3}}}=-{\sqrt {3}}\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {{\frac {3}{t^{2}-1}}+3}}}}=-{\sqrt {3}}\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {3}}{\sqrt {\frac {1+t^{2}-1}{t^{2}-1}}}}}=$

$=-\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {\frac {t^{2}}{t^{2}-1}}}}}=-\int {\frac {t\;dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot t{\sqrt {\frac {1}{t^{2}-1}}}}}=-\int {\frac {dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {\frac {1}{t^{2}-1}}}}}=$

$=-\int {\frac {dt}{\sqrt {\frac {(t^{2}-1)^{3}}{t^{2}-1}}}}=-\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}}}}=-\int {\frac {dt}{t^{2}-1}}=\int {\frac {dt}{1-t^{2}}}.$

Iš interentinio integratoriaus:

$\int {\frac {dt}{1-t^{2}}}={\frac {1}{2}}(\ln(t+1)-\ln(1-t))+C=\tanh ^{-1}t+C={\text{artanh}}\;t+C.$

Kur $t={\sqrt {x^{2}+3 \over x^{2}}}.$ Kad galima būtų skaičiuoti pagal šią formulę t turi būti mažiau už 1 (t<1). Nes kitaip nesiskaičiuoja ln(1-t). Bet mūsų pavyzdyje, kad ir kokias x reikšmes nestatysi į $t={\sqrt {x^{2}+3 \over x^{2}}},$ vistiek t bus daugiau už 1.

Galima taip integruot:

$-\int {\frac {dt}{t^{2}-1}}=-\int {\frac {dt}{(t-1)(t+1)}};$

toliau integruojama kaip racionali funkcija.

${\frac {-1}{(t-1)(t+1)}}={\frac {A}{t-1}}+{\frac {B}{t+1}};$

Abi lygties puses padauginame iš (t-1)(t+1). Tada

$-1=A(t+1)+B(t-1),$

$-1=At+A+Bt-B,$

$-1=(A+B)t+A-B;$

iš čia turime sistemą:

A+B=0,

A-B=-1.

Tada iš antros lygties A=B-1. Įstačius šia A reikšmę į pirmą lygtį, gauname

B-1+B=0,

2B=1,

B=1/2.

Tada A=-B=-1/2.

Tokiu budu gauname, kad

${\frac {-1}{(t-1)(t+1)}}={\frac {-1/2}{t-1}}+{\frac {1/2}{t+1}}.$

Integruodami gauname:

$-\int {\frac {dt}{t^{2}-1}}=-\int {\frac {dt}{(t-1)(t+1)}}=\int ({\frac {-1/2}{t-1}}+{\frac {1/2}{t+1}})dt={\frac {1}{2}}(-\ln(t-1)+\ln(t+1))+C.$

Kur $t={\sqrt {x^{2}+3 \over x^{2}}}.$

1. Jei įstatysime x=2, tai gausime, $t={\sqrt {x^{2}+3 \over x^{2}}}={\sqrt {2^{2}+3 \over 2^{2}}}={\sqrt {7 \over 4}}={\sqrt {1.75}}=1.32287565553.$

${\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln({\sqrt {1.75}}-1)+\ln({\sqrt {1.75}}+1))=0.986646961=$

=0.98664696104483410110205523811797.

Kai x=1, tai $t={\sqrt {x^{2}+3 \over x^{2}}}={\sqrt {1^{2}+3 \over 1^{2}}}={\sqrt {4 \over 1}}={\sqrt {4}}=2.$

${\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln(2-1)+\ln(2+1))={\frac {\ln 3}{2}}=0.549306144334=$

=0.54930614433405484569762261846126.

Bet tokia funkcija integruojama lengviau kitaip (ne per diferencialinius binomus; integruojant keičiant kintamąjį) ir yra jinai integralų lentelėje

$\int {\frac {dx}{\sqrt {x^{2}\pm a^{2}}}}=\ln |x+{\sqrt {x^{2}\pm a^{2}}}|+C.$

2. Tada, kai x=2, tai

$\ln |x+{\sqrt {x^{2}+3}}|=\ln |2+{\sqrt {2^{2}+3}}|=\ln |2+{\sqrt {7}}|=1.5359531=$

=1.5359531053788889467996778565792.

O kai x=1, tai

$\ln |x+{\sqrt {x^{2}+3}}|=\ln |1+{\sqrt {1^{2}+3}}|=\ln |1+{\sqrt {4}}|=\ln |3|=1.0986122886681=$

=1.0986122886681096913952452369225.

Tada, kai x kinta nuo 1 iki 2, tai pirmu atveju integruojant gauname:

0.98664696104483410110205523811797 - 0.54930614433405484569762261846126 = 0.43734081671077925540443261965671.

Antru atveju, kai x kinta nuo 1 iki 2 integruojant gautume:

1.5359531053788889467996778565792 - 1.0986122886681096913952452369225 = 0.4373408167107792554044326196567.

Abiais būdais integruojant gavome tą patį atsakymą.

Toks Free Pascal kodas:

  var a:longint; c,d:real;
  begin
  for a:=1 to 100000000  do
  d:=d+0.00000002/sqrt(sqr(a*0.00000002)+3);
  for a:=1 to 100000000  do
  c:=c+0.00000001/sqrt(sqr(a*0.00000001)+3);
  writeln(d);
  writeln(c);
  writeln(d-c);
  readln;
  end.

Duoda rezultatus:

9.8664695905111544E-001

5.4930614394743249E-001

4.3734081510368294E-001

po 4 sekundžių su 4.16 GHz dažniu veikiančiu procesorium (per pirmus du paleidimus duoda šituos rezultatus po 18 sekundžių; bet jeigu iškart exe failą (diferencialiniaibinomai.exe) paleist [kurį sukuria visada Free Pascal programa] iš "C:\FPC\3.2.0\bin\i386-win32", tai rezultatai gaunami po 4 sekundžių ir taip yra su visais Free Pascal skaičiavimais, kad per exe failą greičiau skaičiuoja [iš pirmo karto]). Matome, kad rezultatai tokie patys kaip skaičiuojant/integruojant pirmu atveju.

Toks Free Pascal kodas:

  var a:longint; c,d:real;
  begin
  for a:=1 to 1000000000  do
  d:=d+0.000000002/sqrt(sqr(a*0.000000002)+3);
  writeln(d);
  readln;
  end.

duoda rezultatą "9.8664696084608883E-001" (tai reiškia $9.8664696084608883\cdot 10^{-1}$) po 20 sekundžių su 4.16 GHz dažniu veikiančiu procesoriumi (per pirmus 2 kartus duodą rezultatą po 34 sekundžių).

Kodas apskaičiuoja plotą, po funkciją $f(x)={\frac {1}{\sqrt {x^{2}+3}}},$ apribotą šios funkcijos kreive, ašimi Ox, ašimi Oy ir ašiai Ox statmena tiese taške x=2.

Beje, integruojant antru budu, kai x kinta nuo 0 iki 2, gauname:

$\int _{0}^{2}{\frac {dx}{\sqrt {x^{2}+3}}}=\ln |x+{\sqrt {x^{2}+3}}||_{0}^{2}=\ln |2+{\sqrt {2^{2}+3}}|-\ln |0+{\sqrt {0^{2}+3}}|=\ln |2+{\sqrt {7}}|-\ln |{\sqrt {3}}|=$

= 1.5359531053788889467996778565792 - 0.54930614433405484569762261846126 = 0.98664696104483410110205523811797.

Į anksčiau pirmu budu gautą integralą, įstatę $t={\sqrt {x^{2}+3 \over x^{2}}}={\frac {\sqrt {x^{2}+3}}{x}},$ gauname:

${\frac {1}{2}}(-\ln(t-1)+\ln(t+1))={\frac {1}{2}}(-\ln({\frac {\sqrt {x^{2}+3}}{x}}-1)+\ln({\frac {\sqrt {x^{2}+3}}{x}}+1))={\frac {1}{2}}\ln {\frac {{\frac {\sqrt {x^{2}+3}}{x}}+1}{{\frac {\sqrt {x^{2}+3}}{x}}-1}}=$

$={\frac {1}{2}}\ln {\frac {\frac {{\sqrt {x^{2}+3}}+x}{x}}{\frac {{\sqrt {x^{2}+3}}-x}{x}}}={\frac {1}{2}}\ln {\frac {{\sqrt {x^{2}+3}}+x}{{\sqrt {x^{2}+3}}-x}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x)).$

Tokiu budu gavome, kad

$\int {\frac {dx}{\sqrt {x^{2}+3}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x))+C.$

O bendru atveju gauname tokį, tikriausiai niekam nematytą, integralą:

$\int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+a^{2}}}-x)+\ln({\sqrt {x^{2}+a^{2}}}+x))+C.$

Integruodami uždavinio sąlygos integralą nuo 0 iki 2, gauname:

$\int _{0}^{2}{\frac {dx}{\sqrt {x^{2}+3}}}={\frac {1}{2}}(-\ln({\sqrt {x^{2}+3}}-x)+\ln({\sqrt {x^{2}+3}}+x))|_{0}^{2}=$

$={\frac {1}{2}}(-\ln({\sqrt {2^{2}+3}}-2)+\ln({\sqrt {2^{2}+3}}+2))-{\frac {1}{2}}(-\ln({\sqrt {0^{2}+3}}-0)+\ln({\sqrt {0^{2}+3}}+0))=$

$={\frac {1}{2}}(-\ln({\sqrt {7}}-2)+\ln({\sqrt {7}}+2))-{\frac {1}{2}}(-\ln {\sqrt {3}}+\ln {\sqrt {3}})={\frac {1}{2}}(-\ln({\sqrt {7}}-2)+\ln({\sqrt {7}}+2))=$

$={\frac {1}{2}}(-(-0.43734081671)+1.535953105)={\frac {1}{2}}\cdot 1.973293922=$

=0.98664696104483410110205523811797.

Toks pat atsakymas, kaip ir integruojant anksčiau.

Dar galima gauti kitokia šio integralo išraišką. Štai taip:

${\frac {1}{2}}\ln {\frac {{\sqrt {x^{2}+3}}+x}{{\sqrt {x^{2}+3}}-x}}={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)({\sqrt {x^{2}+3}}+x)}{({\sqrt {x^{2}+3}}-x)({\sqrt {x^{2}+3}}+x)}}={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)^{2}}{(x^{2}+3)-x^{2}}}=$

$={\frac {1}{2}}\ln {\frac {({\sqrt {x^{2}+3}}+x)^{2}}{3}}=-{\frac {1}{2}}\ln |3|+{\frac {1}{2}}\ln |({\sqrt {x^{2}+3}}+x)^{2}|=-{\frac {1}{2}}\ln |3|+\ln |{\sqrt {x^{2}+3}}+x|.$

Tada $\int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}=-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|+C.$

Integruodami iš pirmo būdo ką tik gautą integralą nuo 0 iki x, turėsime:

$\int _{0}^{x}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}=(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)|_{0}^{x}=(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)-(-\ln |a|+\ln |{\sqrt {0^{2}+a^{2}}}+0|)=$

$=(-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|)-(-\ln |a|+\ln |{\sqrt {a^{2}}}|)=-\ln |a|+\ln |{\sqrt {x^{2}+a^{2}}}+x|.$

Antru budu integruodami nuo 0 iki x, turime:

$\int _{0}^{x}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}=\ln |x+{\sqrt {x^{2}+a^{2}}}||_{0}^{x}=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |0+{\sqrt {0^{2}+a^{2}}}|=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |{\sqrt {a^{2}}}|=\ln |x+{\sqrt {x^{2}+a^{2}}}|-\ln |a|.$

Gavome tokius pačius integralus ir įstačius vietoj x ir a bet kokias reikšmes, abiais būdais gausime tokias pat išraiškas ir atsakymus.

• $\int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx;$

$m=-{\frac {1}{2}},\;\;n={\frac {1}{4}},\;\;p={\frac {1}{3}},\;\;{\frac {m+1}{n}}={\frac {-1/2+1}{1/4}}={\frac {1/2}{1/4}}=2\;$ (II atvejis).

Keičiame $a+bx^{n}=t^{\alpha },$ kur $\alpha$ - trupmenos p vardiklis. Tada pakeitimas yra $1+x^{1/4}=t^{3},\;$ $x^{1/4}=t^{3}-1,\;$ $x=(t^{3}-1)^{4},\;$

$dx=4(t^{3}-1)^{3}\cdot 3t^{2}\;dt=12t^{2}(t^{3}-1)^{3}\;dt.\;$ ${\sqrt[{3}]{1+x^{1/4}}}=t.$

$\int {\frac {\sqrt[{3}]{1+{\sqrt[{4}]{x}}}}{\sqrt {x}}}dx=\int x^{-1/2}(1+x^{1/4})^{1/3}dx=\int (t^{3}-1)^{-2}(1+(t^{3}-1))^{1/3}12t^{2}(t^{3}-1)^{3}\;dt=$

$=\int (1+(t^{3}-1))^{1/3}12t^{2}(t^{3}-1)\;dt=12\int (t^{3})^{1/3}t^{2}(t^{3}-1)\;dt=12\int t^{3}(t^{3}-1)\;dt=12\int (t^{6}-t^{3})\;dt=$

$=12({\frac {t^{7}}{7}}-{\frac {t^{4}}{4}})+C={\frac {3}{7}}t^{4}(4t^{3}-7)+C={\frac {3}{7}}({\sqrt[{3}]{1+x^{1/4}}})^{4}(4(1+x^{1/4})-7)+C={\frac {3}{7}}{\sqrt[{3}]{1+x^{1/4}}}(1+x^{1/4})(4(1+x^{1/4})-7)+C.$

• Apskaičiuosime integralą

$I=\int {\frac {dx}{x^{2}{\sqrt {a+bx^{2}}}}}=\int x^{-2}(a+bx^{2})^{-{\frac {1}{2}}}dx.$

Šiame pavyzdyje $m=-2,\;n=2,\;p=-{\frac {1}{2}},$ todėl ${\frac {m+1}{n}}+p={\frac {-2+1}{2}}-{\frac {1}{2}}={\frac {-1}{2}}-{\frac {1}{2}}=-1\;$ (III atvejis).

Tada $a+bx^{n}=t^{\alpha }x^{n},$ kur $\alpha$ - trupmenos p vardiklis; $a+bx^{2}=t^{2}x^{2}.$ Ir pasinaudojame keitiniu

$t={\sqrt {{\frac {a}{x^{2}}}+b}};\;\;t^{2}x^{2}-bx^{2}=a,\;(t^{2}-b)x^{2}=a,\;x={\frac {\sqrt {a}}{\sqrt {t^{2}-b}}};\;\;dx=-{\frac {1}{2}}{\frac {{\sqrt {a}}\;2t\;dt}{(t^{2}-b)^{3/2}}}={\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}.$

Tada gauname

$I=\int x^{-2}(a+bx^{2})^{-{\frac {1}{2}}}dx=\int {\frac {t^{2}-b}{a}}(a+b{\frac {a}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=\int {\frac {t^{2}-b}{a}}({\frac {a(t^{2}-b)+ab}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=$

$=\int {\frac {t^{2}-b}{a}}({\frac {t^{2}}{t^{2}-b}})^{-{\frac {1}{2}}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=\int {\frac {t^{2}-b}{a}}{\frac {\sqrt {t^{2}-b}}{t}}{\frac {-{\sqrt {a}}\;t\;dt}{\sqrt {(t^{2}-b)^{3}}}}=-\int {\frac {dt}{\sqrt {a}}}=$

$=-{\frac {t}{\sqrt {a}}}+C=-{\frac {\sqrt {{\frac {a}{x^{2}}}+b}}{\sqrt {a}}}+C.$

• Apskaičiuosime integralą $I=\int x^{5}(1-x^{2})^{-{\frac {1}{2}}}dx.$ Šiuo atveju $m=5,\;n=2,\;p=-{\frac {1}{2}},$ todėl ${\frac {m+1}{n}}=3\;$ (II atvejis). Pasinaudoję pakeitimais

$t={\sqrt {1-x^{2}}},\;t^{2}=1-x^{2},\;x={\sqrt {1-t^{2}}},\;\;dx={\frac {-2t\;dt}{2{\sqrt {1-t^{2}}}}}={\frac {-t\;dt}{\sqrt {1-t^{2}}}},$

gauname

$I=\int x^{5}(1-x^{2})^{-{\frac {1}{2}}}dx=\int ({\sqrt {1-t^{2}}})^{5}(1-[1-t^{2}])^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=\int {\sqrt {1-t^{2}}}(1-t^{2})^{2}(t^{2})^{-{\frac {1}{2}}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=$

$=\int {\sqrt {1-t^{2}}}(1-t^{2})^{2}{\frac {1}{t}}{\frac {-t\;dt}{\sqrt {1-t^{2}}}}=-\int (1-t^{2})^{2}\;dt=-\int (1-2t^{2}+t^{4})dt=-\int dt+2\int t^{2}\;dt-t^{4}\;dt=$

$=-t+{\frac {2}{3}}t^{3}-{\frac {t^{5}}{5}}+C=-{\sqrt {1-x^{2}}}+{\frac {2}{3}}{\sqrt {(1-x^{2})^{3}}}-{\frac {\sqrt {(1-x^{2})^{5}}}{5}}+C.$

Nuorodos

• Oilerio keitiniai
• https://en.wikipedia.org/wiki/Euler_substitution Oilerio keitiniai