∫ d x x 2 + 3 = ∫ t 2 + 3 2 t 2 d t t 2 + 3 2 t = ∫ d t t = ln | t | + C = ln | x + x 2 + 3 | + C . {\displaystyle \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.}
Patikriname ar atsakymai bus vienodi istacius x=4.
1 x 2 + 3 = 1 4 2 + 3 = 1 19 = 1 4.358898944 = 0.229415733. {\displaystyle {\frac {1}{\sqrt {x^{2}+3}}}={\frac {1}{\sqrt {4^{2}+3}}}={\frac {1}{\sqrt {19}}}={\frac {1}{4.358898944}}=0.229415733.} ( ln | x + x 2 + 3 | + C ) ′ = ( x + x 2 + 3 ) ′ x + x 2 + 3 + C ′ = 1 + 1 2 ⋅ ( x 2 + 3 ) ′ x 2 + 3 x + x 2 + 3 = 1 + 2 x 2 x 2 + 3 x + x 2 + 3 = 1 + x x 2 + 3 x + x 2 + 3 = {\displaystyle (\ln |x+{\sqrt {x^{2}+3}}|+C)'={\frac {(x+{\sqrt {x^{2}+3}})'}{x+{\sqrt {x^{2}+3}}}}+C'={\frac {1+{1 \over 2}\cdot {\frac {(x^{2}+3)'}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {2x}{2{\sqrt {x^{2}+3}}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {x}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}=}
= 1 + 4 4 2 + 3 4 + 4 2 + 3 = 1 + 4 19 4 + 19 = 1 + 4 4.358898944 4 + 4.358898944 = 1 + 0.917662935 8.358898944 = 1.917662935 8.358898944 = 0.229415733. {\displaystyle ={\frac {1+{\frac {4}{\sqrt {4^{2}+3}}}}{4+{\sqrt {4^{2}+3}}}}={\frac {1+{\frac {4}{\sqrt {19}}}}{4+{\sqrt {19}}}}={\frac {1+{\frac {4}{4.358898944}}}{4+4.358898944}}={\frac {1+0.917662935}{8.358898944}}={\frac {1.917662935}{8.358898944}}=0.229415733.}
∫ x d x ( 7 x − 10 − x 2 ) 3 = ∫ 5 + 2 t 2 1 + t 2 − 6 t ( 1 + t 2 ) 2 d t ( 3 t 1 + t 2 ) 3 = ∫ − 6 t ( 5 + 2 t 2 ) ( 1 + t 2 ) 3 ( 3 t 1 + t 2 ) 3 d t = ∫ − 6 t ( 5 + 2 t 2 ) 27 t 3 d t = ∫ − 6 ( 5 + 2 t 2 ) 27 t 2 d t = {\displaystyle \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=}
= ∫ ( − 30 27 t 2 − 12 t 2 27 t 2 ) d t = 1 27 ∫ ( − 30 t 2 − 12 ) d t = 1 27 ( 30 t − 12 t ) + C = 1 27 ( 30 ( 5 − x ) ( x − 2 ) − 12 ( 5 − x ) ( x − 2 ) ) + C = {\displaystyle =\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=}
= 30 x − 2 27 5 − x − 12 27 ( 5 − x ) ( x − 2 ) + C . {\displaystyle ={\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {(5-x)}{(x-2)}}}+C.}
Patikriname ar atsakymai bus vienodi istacius x = 3 {\displaystyle x=3} .
( 30 x − 2 27 5 − x − 12 27 5 − x x − 2 + C ) ′ = 30 27 ⋅ 1 2 ⋅ 5 − x x − 2 ⋅ ( x − 2 5 − x ) ′ − 12 27 ⋅ 1 2 ⋅ x − 2 5 − x ⋅ ( 5 − x x − 2 ) ′ = {\displaystyle ({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {1}{2}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot ({\frac {x-2}{5-x}})'-{\frac {12}{27}}\cdot {\frac {1}{2}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot ({\frac {5-x}{x-2}})'=} = 15 27 ⋅ 5 − x x − 2 ⋅ x ( 5 − x ) − ( x − 2 ) ⋅ ( − x ) ( 5 − x ) 2 − 6 27 ⋅ x − 2 5 − x ⋅ − x ( x − 2 ) − ( 5 − x ) x ( x − 2 ) 2 = {\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {x(5-x)-(x-2)\cdot (-x)}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x(x-2)-(5-x)x}{(x-2)^{2}}}=}
= 15 27 ⋅ 5 − x x − 2 ⋅ 5 x − x 2 + x 2 − 2 x ( 5 − x ) 2 − 6 27 ⋅ x − 2 5 − x ⋅ − x 2 + 2 x − 5 x + x 2 ( x − 2 ) 2 = {\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {5x-x^{2}+x^{2}-2x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x^{2}+2x-5x+x^{2}}{(x-2)^{2}}}=}
= 15 27 ⋅ 5 − x x − 2 ⋅ 3 x ( 5 − x ) 2 − 6 27 ⋅ x − 2 5 − x ⋅ − 3 x ( x − 2 ) 2 = 1 27 ( 5 − 3 3 − 2 ⋅ 15 ⋅ 3 ⋅ 3 ( 5 − 3 ) 2 + 3 − 2 5 − 3 ⋅ 18 ⋅ 3 ( 3 − 2 ) 2 ) = {\displaystyle ={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {3x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-3x}{(x-2)^{2}}}={\frac {1}{27}}({\sqrt {\frac {5-3}{3-2}}}\cdot {\frac {15\cdot 3\cdot 3}{(5-3)^{2}}}+{\frac {\sqrt {3-2}}{\sqrt {5-3}}}\cdot {\frac {18\cdot 3}{(3-2)^{2}}})=}
= 1 27 ( 2 1 ⋅ 135 2 2 + 1 2 ⋅ 54 1 2 ) = 1 27 ( 135 2 4 + 54 2 ) = 1 27 ( 47.72970773 + 38.18376618 ) = 3.181980515. {\displaystyle ={\frac {1}{27}}({\sqrt {\frac {2}{1}}}\cdot {\frac {135}{2^{2}}}+{\frac {\sqrt {1}}{\sqrt {2}}}\cdot {\frac {54}{1^{2}}})={\frac {1}{27}}({\frac {135{\sqrt {2}}}{4}}+{\frac {54}{\sqrt {2}}})={\frac {1}{27}}(47.72970773+38.18376618)=3.181980515.}
x ( 7 x − 10 − x 2 ) 3 = 3 ( 7 ⋅ 3 − 10 − 3 2 ) 3 = 3 ( 21 − 10 − 9 ) 3 = 3 2 3 = 3 8 = 1.060660172. {\displaystyle {\frac {x}{\sqrt {(7x-10-x^{2})^{3}}}}={\frac {3}{\sqrt {(7\cdot 3-10-3^{2})^{3}}}}={\frac {3}{\sqrt {(21-10-9)^{3}}}}={\frac {3}{\sqrt {2^{3}}}}={\frac {3}{\sqrt {8}}}=1.060660172.} Pabandome gauti teisinga atsakyma isvestine darant kitokiu budu (ir istacius x=3), nors sis budas ligtais 95%, kad neteisingas:
( 30 x − 2 27 5 − x − 12 27 5 − x x − 2 + C ) ′ = 30 27 ⋅ 5 − x 2 x − 2 − x − 2 2 5 − x ( 5 − x ) 2 − 12 27 ⋅ x − 2 2 5 − x − 5 − x 2 x − 2 ( x − 2 ) 2 = {\displaystyle ({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}-{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}}{({\sqrt {5-x}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}-{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}}{({\sqrt {x-2}})^{2}}}=} = 30 27 ⋅ 5 − 3 2 3 − 2 − 3 − 2 2 5 − 3 ( 5 − 3 ) 2 − 12 27 ⋅ 3 − 2 2 5 − 3 − 5 − 3 2 3 − 2 ( 3 − 2 ) 2 = 30 27 ⋅ 2 2 1 − 1 2 2 ( 2 ) 2 − 12 27 ⋅ 1 2 2 − 2 2 1 ( 1 ) 2 = {\displaystyle ={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}-{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}}{({\sqrt {5-3}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}-{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}}{({\sqrt {3-2}})^{2}}}={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {2}}{2{\sqrt {1}}}}-{\frac {\sqrt {1}}{2{\sqrt {2}}}}}{({\sqrt {2}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {1}}{2{\sqrt {2}}}}-{\frac {\sqrt {2}}{2{\sqrt {1}}}}}{({\sqrt {1}})^{2}}}=}
= 30 27 ⋅ 2 − 1 2 2 2 − 12 27 ⋅ 1 − 2 2 2 = 30 27 ⋅ 1 4 2 + 12 27 ⋅ 1 2 2 = 15 54 2 + 6 27 2 = 15 − 12 54 2 = 3 54 2 = 1 18 2 = 0.03928371. {\displaystyle ={\frac {30}{27}}\cdot {\frac {\frac {2-1}{2{\sqrt {2}}}}{2}}-{\frac {12}{27}}\cdot {\frac {1-2}{2{\sqrt {2}}}}={\frac {30}{27}}\cdot {\frac {1}{4{\sqrt {2}}}}+{\frac {12}{27}}\cdot {\frac {1}{2{\sqrt {2}}}}={\frac {15}{54{\sqrt {2}}}}+{\frac {6}{27{\sqrt {2}}}}={\frac {15-12}{54{\sqrt {2}}}}={\frac {3}{54{\sqrt {2}}}}={\frac {1}{18{\sqrt {2}}}}=0.03928371.}
Pabandome gauti teisinga atsakyma isvestine darant kitokiu budu (ir istacius x=3), nors sis budas ligtais 95%, kad neteisingas:
( 30 x − 2 27 5 − x − 12 27 5 − x x − 2 + C ) ′ = {\displaystyle ({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'=} = 30 27 ( 1 2 ⋅ ( x − 2 ) ′ x − 2 5 − x + − x − 2 ⋅ ( 5 − x ) ′ ( 5 − x ) 2 ) − 12 27 ( 1 2 ⋅ ( 5 − x ) ′ 5 − x x − 2 + − 5 − x ⋅ ( x − 2 ) ′ ( x − 2 ) 2 ) = {\displaystyle ={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {(x-2)'}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot ({\sqrt {5-x}})'}{({\sqrt {5-x}})^{2}}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {(5-x)'}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot ({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}})=}
= 30 27 ( 1 2 ⋅ 1 x − 2 5 − x + − x − 2 ⋅ 1 2 ⋅ ( 5 − x ) ′ 5 − x ⋅ ( 5 − x ) ) − 12 27 ( 1 2 ⋅ − 1 5 − x x − 2 + − 5 − x ⋅ 1 2 ( x − 2 ) ′ x − 2 ( x − 2 ) ) = {\displaystyle ={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot {\frac {1}{2}}\cdot (5-x)'}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot {\frac {1}{2}}(x-2)'}{{\sqrt {x-2}}(x-2)}})=}
= 15 27 ( 1 x − 2 5 − x + − x − 2 ⋅ ( − 1 ) 5 − x ⋅ ( 5 − x ) ) − 6 27 ( − 1 5 − x x − 2 + − 5 − x ⋅ 1 x − 2 ( x − 2 ) ) = {\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot (-1)}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {6}{27}}({\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot 1}{{\sqrt {x-2}}(x-2)}})=}
= 15 27 ( 1 x − 2 5 − x + x − 2 5 − x ⋅ ( 5 − x ) ) + 6 27 ( 1 5 − x x − 2 + 5 − x x − 2 ( x − 2 ) ) = {\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {\sqrt {x-2}}{{\sqrt {5-x}}\cdot (5-x)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {\sqrt {5-x}}{{\sqrt {x-2}}(x-2)}})=}
= 15 27 ( 1 3 − 2 5 − 3 + 3 − 2 5 − 3 ⋅ ( 5 − 3 ) ) + 6 27 ( 1 5 − 3 3 − 2 + 5 − 3 3 − 2 ( 3 − 2 ) ) = {\displaystyle ={\frac {15}{27}}({\frac {1}{{\sqrt {3-2}}{\sqrt {5-3}}}}+{\frac {\sqrt {3-2}}{{\sqrt {5-3}}\cdot (5-3)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-3}}{\sqrt {3-2}}}}+{\frac {\sqrt {5-3}}{{\sqrt {3-2}}(3-2)}})=}
= 15 27 ( 1 2 + 1 2 ⋅ 2 ) + 6 27 ( 1 2 + 2 1 ) = 15 27 ⋅ 2 + 1 2 2 + 6 27 ⋅ 1 + 2 2 = 45 54 2 + 18 27 2 = 5 6 2 + 2 3 2 = {\displaystyle ={\frac {15}{27}}({\frac {1}{\sqrt {2}}}+{\frac {1}{{\sqrt {2}}\cdot 2}})+{\frac {6}{27}}({\frac {1}{\sqrt {2}}}+{\frac {\sqrt {2}}{1}})={\frac {15}{27}}\cdot {\frac {2+1}{2{\sqrt {2}}}}+{\frac {6}{27}}\cdot {\frac {1+2}{\sqrt {2}}}={\frac {45}{54{\sqrt {2}}}}+{\frac {18}{27{\sqrt {2}}}}={\frac {5}{6{\sqrt {2}}}}+{\frac {2}{3{\sqrt {2}}}}=}
= 0.589255651 + 0.47140452 = 1.060660172. {\displaystyle =0.589255651+0.47140452=1.060660172.}
Sis budas pasirode esas teisingas. Cia buvo pasinaudota dviejomis formulemis:
( f g ) ′ = f ′ g + f g ′ ; {\displaystyle \left({fg}\right)'=f'g+fg';}
( 1 f ) ′ = − f ′ f 2 , f ≠ 0. {\displaystyle \left({\frac {1}{f}}\right)'={\frac {-f'}{f^{2}}},\qquad f\neq 0.}
Diferencijavimas vyko tokiu budu, kad pavyzdiui, sioje funkcijoje 5 − x x − 2 {\displaystyle {\sqrt {\frac {5-x}{x-2}}}} yra dvi funkcijos f ( x ) = f = 5 − x {\displaystyle f(x)=f={\sqrt {5-x}}} , o g ( x ) = g = 1 x − 2 {\displaystyle g(x)=g={\frac {1}{\sqrt {x-2}}}} . Tuomet ( f ( x ) ) ′ = f ′ = 1 2 5 − x {\displaystyle (f(x))'=f'={\frac {1}{2{\sqrt {5-x}}}}} , o ( g ( x ) ) ′ = g ′ = − ( x − 2 ) ′ ( x − 2 ) 2 = − 1 2 x − 2 ( x − 2 ) 2 . {\displaystyle (g(x))'=g'={\frac {-({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}}={\frac {-1}{2{\sqrt {x-2}}({\sqrt {x-2}})^{2}}}.}
Diferencialinių binomų integravimas
Integralas ∫ x m ( a + b x n ) p d x , {\displaystyle \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 , {\displaystyle p>0,} tai pointegralinis binomas skleidžiamas pagal Niutono binomo formulę . Jei p < 0 , {\displaystyle p<0,} tai keičiame x = t k , {\displaystyle x=t^{k},} kur k - bendras trupmenų m ir n vardiklis. Pavyzdžiui, trupmenų 1 4 {\displaystyle {\tfrac {1}{4}}} ir 2 3 {\displaystyle {\tfrac {2}{3}}} bendras vardiklis yra 3 ⋅ {\displaystyle \cdot } 4 = 12.
II. m + 1 n {\displaystyle {\frac {m+1}{n}}} - sveikasis skaičius. Keičiame a + b x n = t α , {\displaystyle a+bx^{n}=t^{\alpha },} kur α {\displaystyle \alpha } - trupmenos p vardiklis.
III. m + 1 n + p {\displaystyle {\frac {m+1}{n}}+p} - sveikasis skaičius. Keičiame a + b x n = t α x n , {\displaystyle a+bx^{n}=t^{\alpha }x^{n},} kur α {\displaystyle \alpha } - trupmenos p vardiklis. Pavyzdžiai
∫ d x x 2 + 3 . {\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}.} Matome, kad tinka trečias atvejis, nes m + 1 n + p = 0 + 1 2 − 1 2 = 0 {\displaystyle {\frac {m+1}{n}}+p={\frac {0+1}{2}}-{\frac {1}{2}}=0} . Čia m=0, n=2, p = 1 2 {\displaystyle p={\frac {1}{2}}} . Keičiame a + b x n = t α x n , {\displaystyle a+bx^{n}=t^{\alpha }x^{n},} kur α {\displaystyle \alpha } - trupmenos p vardiklis. Taigi a + b x n = t α x n {\displaystyle a+bx^{n}=t^{\alpha }x^{n}} , čia a=3, b=1; x 2 + 3 = t 2 x 2 {\displaystyle x^{2}+3=t^{2}x^{2}} ; 3 = t 2 x 2 − x 2 {\displaystyle 3=t^{2}x^{2}-x^{2}} ; x 2 = 3 t 2 − 1 = 3 ( t 2 − 1 ) − 1 {\displaystyle x^{2}={\frac {3}{t^{2}-1}}=3(t^{2}-1)^{-1}} ; x = 3 ⋅ ( t 2 − 1 ) − 1 2 {\displaystyle x={\sqrt {3}}\cdot (t^{2}-1)^{-{\frac {1}{2}}}} .d x = − 3 2 ⋅ ( t 2 − 1 ) − 3 2 d t = − 3 2 ( t 2 − 1 ) 3 d t . {\displaystyle dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}dt=-{\frac {\sqrt {3}}{2{\sqrt {(t^{2}-1)^{3}}}}}dt.}
∫ d x x 2 + 3 = − 3 2 ∫ d t ( t 2 − 1 ) 3 ⋅ 3 t 2 − 1 + 3 = − 3 2 3 ∫ d t ( t 2 − 1 ) 2 + ( t 2 − 1 ) 3 = {\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}=-{\frac {\sqrt {3}}{2}}\int {\frac {dt}{{\sqrt {(t^{2}-1)^{3}}}\cdot {\sqrt {{\frac {3}{t^{2}-1}}+3}}}}=-{\frac {\sqrt {3}}{2{\sqrt {3}}}}\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=}
= − 1 2 ∫ d t ( t 4 − 2 t 2 + 1 ) + ( t 6 − 3 ( t 2 ) 2 + 3 t 2 − 1 3 ) = − 1 2 ∫ d t t 6 − 2 t 4 + t 2 = {\displaystyle =-{\frac {1}{2}}\int {\frac {dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-{\frac {1}{2}}\int {\frac {dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=}
= − 1 2 ∫ d t t t 4 − 2 t 2 + 1 = − 1 2 ∫ d t t ( t 2 − 1 ) 2 = − 1 2 ∫ d t t ( t 2 − 1 ) = − 1 2 ∫ ( t t 2 − 1 − 1 t ) d t = {\displaystyle =-{\frac {1}{2}}\int {\frac {dt}{t{\sqrt {t^{4}-2t^{2}+1}}}}=-{\frac {1}{2}}\int {\frac {dt}{t{\sqrt {(t^{2}-1)^{2}}}}}=-{\frac {1}{2}}\int {\frac {dt}{t(t^{2}-1)}}=-{\frac {1}{2}}\int {\Big (}{\frac {t}{t^{2}-1}}-{\frac {1}{t}}{\Big )}dt=} = − 1 4 ( ln ( t 2 − 1 ) − 2 ln t ) . {\displaystyle =-{1 \over 4}(\ln(t^{2}-1)-2\ln t).} Kur t = x 2 + 3 x 2 . {\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.}
Iš tikro d x = − 3 2 ⋅ ( t 2 − 1 ) − 3 2 2 t d t = − 3 ( t 2 − 1 ) 3 t d t . {\displaystyle dx=-{\frac {\sqrt {3}}{2}}\cdot (t^{2}-1)^{-{\frac {3}{2}}}2t\;dt=-{\frac {\sqrt {3}}{\sqrt {(t^{2}-1)^{3}}}}t\;dt.}
∫ d x x 2 + 3 . {\displaystyle \int {\frac {dx}{\sqrt {x^{2}+3}}}.} Matome, kad tinka trečias atvejis, nes m + 1 n + p = 0 + 1 2 − 1 2 = 0 {\displaystyle {\frac {m+1}{n}}+p={\frac {0+1}{2}}-{\frac {1}{2}}=0} . Čia m=0, n=2, p = 1 2 {\displaystyle p={\frac {1}{2}}} . Keičiame a + b x n = t α x n , {\displaystyle a+bx^{n}=t^{\alpha }x^{n},} kur α {\displaystyle \alpha } - trupmenos p vardiklis. Taigi a + b x n = t α x n {\displaystyle a+bx^{n}=t^{\alpha }x^{n}} , čia a=3, b=1; x 2 + 3 = t 2 x 2 {\displaystyle x^{2}+3=t^{2}x^{2}} ; 3 = t 2 x 2 − x 2 {\displaystyle 3=t^{2}x^{2}-x^{2}} ; x 2 = 3 t 2 − 1 = 3 ( t 2 − 1 ) − 1 {\displaystyle x^{2}={\frac {3}{t^{2}-1}}=3(t^{2}-1)^{-1}} ; x = 3 ⋅ ( t 2 − 1 ) − 1 2 {\displaystyle x={\sqrt {3}}\cdot (t^{2}-1)^{-{\frac {1}{2}}}} .d x = − 3 2 ⋅ ( t 2 − 1 ) − 3 2 ⋅ 2 t d t = − 3 t ( t 2 − 1 ) 3 d t . {\displaystyle 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.}
∫ d x x 2 + 3 = − 3 ∫ t d t ( t 2 − 1 ) 3 ⋅ 3 t 2 − 1 + 3 = − 3 3 ∫ t d t ( t 2 − 1 ) 2 + ( t 2 − 1 ) 3 = {\displaystyle \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}}}}=-{\frac {\sqrt {3}}{\sqrt {3}}}\int {\frac {t\;dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=}
= − ∫ t d t ( t 4 − 2 t 2 + 1 ) + ( t 6 − 3 ( t 2 ) 2 + 3 t 2 − 1 3 ) = − ∫ t d t t 6 − 2 t 4 + t 2 = {\displaystyle =-\int {\frac {t\;dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-\int {\frac {t\;dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=}
= − ∫ t d t t t 4 − 2 t 2 + 1 = − ∫ d t ( t 2 − 1 ) 2 = − ∫ d t t 2 − 1 = − ∫ d t ( t − 1 ) ( t + 1 ) ; {\displaystyle =-\int {\frac {t\;dt}{t{\sqrt {t^{4}-2t^{2}+1}}}}=-\int {\frac {dt}{\sqrt {(t^{2}-1)^{2}}}}=-\int {\frac {dt}{t^{2}-1}}=-\int {\frac {dt}{(t-1)(t+1)}};}
toliau integruojama kaip racionali funkcija.
− 1 ( t − 1 ) ( t + 1 ) = A t − 1 + B t + 1 ; {\displaystyle {\frac {-1}{(t-1)(t+1)}}={\frac {A}{t-1}}+{\frac {B}{t+1}};}
Kur t = x 2 + 3 x 2 . {\displaystyle t={\sqrt {x^{2}+3 \over x^{2}}}.} Bet tokia funkcija integruojama lengviau kitaip (ne per diferencialinius binomus) ir yra jinai integralų lentelėje
∫ d x x 2 ± a 2 = ln | x + x 2 ± a 2 | + C . {\displaystyle \int {\frac {dx}{\sqrt {x^{2}\pm a^{2}}}}=\ln |x+{\sqrt {x^{2}\pm a^{2}}}|+C.} Klaida čia:
∫ d x x 2 + 3 = − 3 ∫ t d t ( t 2 − 1 ) 3 ⋅ 3 t 2 − 1 + 3 = − 3 3 ∫ t d t ( t 2 − 1 ) 2 + ( t 2 − 1 ) 3 = {\displaystyle \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}}}}=-{\frac {\sqrt {3}}{\sqrt {3}}}\int {\frac {t\;dt}{\sqrt {(t^{2}-1)^{2}+(t^{2}-1)^{3}}}}=} Paraboloid (aptarimas ) 19:55, 31 gruodžio 2022 (UTC) Atsakyti
Pasirodo, kad teisingai skaičiuojant gaunamas toks pat integralas − ∫ d t t 2 − 1 . {\displaystyle -\int {\frac {dt}{t^{2}-1}}.} Tik dabar niekaip negaliu suprasti kaip buvo priskaičiuota teisingai, šiuo iš pažiūros klaidingu budu, kuris yra kažkaip teisingas. Arba ten tik galutinis atsakymas (− ∫ d t t 2 − 1 {\displaystyle -\int {\frac {dt}{t^{2}-1}}} ) teisingas, o ne pats skaičiavimo būdas...
Dabar atrodo supratau, kaip ten skaičiuota (ir ten teisingai skaičiuota)...