Aptarimas:Matematika/Iracionaliųjų funkcijų integravimas

Pirmi du variantai galetu buti isspresti kitaip - taip kaip atrodytu nenusizengiant taisyklem integravimo ir be prikabinimo is nieko kitu demenu. Atsakymai zinoma skirtusi. Stai ju sprendimo budai:

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

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

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

$=-{\frac {\sqrt {5}}{5}}\ln |1-2{\sqrt {5}}u|+C=-{\frac {\sqrt {5}}{5}}\ln |1-2{\sqrt {5}}{\sqrt {5x^{2}+x+1}}|+C=$ $=-{\frac {\sqrt {5}}{5}}\ln |1-2{\sqrt {25x^{2}+5x+5}}|+C.$

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

${\sqrt {1-6x-9x^{2}}}=ux+1.$ 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: $-9dx=2uxdu+2du;$ $dx={\frac {-2(ux+1)}{9}}du$ ; $u={\frac {{\sqrt {1-6x-9x^{2}}}-1}{x}}.$

$\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=-2\int {\frac {(ux+1)du}{9(ux+1)}}=-2\int {\frac {du}{9}}=-{\frac {2}{9}}u+C=$

$=-{\frac {2}{9}}(1-6x-9x^{2})+C.$

Ar Oi-lerio keitiniai nereikalauja papildomu sprendimo ziniu ar salygu, o gal is vis yra klaidingi, tai del to kyla abejoniu del ju tinkamumo matematikoje. Ne veltui vadinasi Oiiii-liar'io keitiniai.

pavyzdio problema

Is straipsnio:

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

Pavyzdyje si vieta $2-x={\frac {4t^{3}}{1+t^{3}}};$ $dx={\frac {-12t^{2}}{(1+t^{3})^{2}}}dt$ atrodo, kad isvestine gauta blogai. Nes

Dalybos taisyklė

$\left({f \over g}\right)'={f'g-fg' \over g^{2}},\qquad g\neq 0$

Tada $f=4t^{3}$, $g=1+t^{3}$. Tada $-dx={12t^{2}(1+t^{3})-4t^{3}\cdot 3t^{2} \over (1+t^{3})^{2}}dt={12t^{2}+12t^{5}-12t^{5} \over (1+t^{3})^{2}}dt={12t^{2} \over (1+t^{3})^{2}};$

isvestines gavimo budas

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

$\left({f \over g}\right)'={f'g-fg' \over g^{2}},\qquad g\neq 0$

$x={\frac {u^{2}-1}{1-2{\sqrt {5}}u}};$

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

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

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

Palyginus su sita nesamone:

$dx=d(x+1)=2udu+2{\sqrt {5}}xdu+2{\sqrt {5}}udx;\;dx={\frac {2(u+{\sqrt {5}}x)}{1-2{\sqrt {5}}u}}du.$

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

$=-{\frac {\sqrt {5}}{5}}\ln |1-2{\sqrt {5}}u|+C=-{\frac {\sqrt {5}}{5}}\ln |1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x)|+C=$ $=-{\frac {\sqrt {5}}{5}}\ln |1+10x-2{\sqrt {25x^{2}+5x+5}}|+C.$

bandymas isspresti pavyzdi normaliau

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

${\sqrt {1-6x-9x^{2}}}=ux+1.$ Pakelę abi puses kvadratu, gauname $1-6x-9x^{2}=u^{2}x^{2}+2ux+1;$ $-6x-9x^{2}=u^{2}x^{2}+2ux;$ $-6-9x=u^{2}x+2u;$ $u^{2}x+9x=-6-2u$; $10x=-6-2u$; $x={\frac {-6-2u}{10}}={\frac {-6}{10}}-{\frac {u}{5}}$ ; $2u=-10x-6$; $u=-5x-3$ ; ${\sqrt {1-6x-9x^{2}}}=u\cdot {\frac {-6-2u}{10}}+1.$ Imdami apiejų lygybės pusių diferencialus, randame: $dx=-{\frac {du}{5}}.$

$\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=-{\frac {1}{5}}\int {\frac {du}{u\cdot {\frac {-6-2u}{10}}+1}}=-{\frac {1}{5}}\int {\frac {du}{-{\frac {3u}{5}}-{\frac {u^{2}}{5}}+1}}=-{\frac {1}{5}}\int {\frac {du}{\frac {-3u-u^{2}+5}{5}}}=$

$=-\int {\frac {du}{-3u-u^{2}+5}}=\int {\frac {du}{u^{2}+3u-5}}=$

For $a\neq 0:$

$\int {\frac {1}{ax^{2}+bx+c}}dx={\begin{cases}\displaystyle {\frac {2}{\sqrt {4ac-b^{2}}}}\arctan {\frac {2ax+b}{\sqrt {4ac-b^{2}}}}+C&{\mbox{(for }}4ac-b^{2}>0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{\sqrt {b^{2}-4ac}}}\,\mathrm {arctanh} {\frac {2ax+b}{\sqrt {b^{2}-4ac}}}+C={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C&{\mbox{(for }}4ac-b^{2}<0{\mbox{)}}\\[12pt]\displaystyle -{\frac {2}{2ax+b}}+C&{\mbox{(for }}4ac-b^{2}=0{\mbox{)}}\end{cases}}$

Mums tinka antras variantas:

$\int {\frac {1}{ax^{2}+bx+c}}dx={\frac {1}{\sqrt {b^{2}-4ac}}}\ln \left|{\frac {2ax+b-{\sqrt {b^{2}-4ac}}}{2ax+b+{\sqrt {b^{2}-4ac}}}}\right|+C$

$\int {\frac {dx}{\sqrt {1-6x-9x^{2}}}}=\int {\frac {du}{u^{2}+3u-5}}={\frac {1}{\sqrt {3^{2}-4\cdot (-5)}}}\ln \left|{\frac {2u+3-{\sqrt {3^{2}+20}}}{2u+3+{\sqrt {9+20}}}}\right|+C={\frac {1}{\sqrt {29}}}\ln \left|{\frac {2u+3-{\sqrt {29}}}{2u+3+{\sqrt {29}}}}\right|+C=$

$={\frac {1}{\sqrt {29}}}\ln \left|{\frac {2(-5x-3)+3-{\sqrt {29}}}{2(-5x-3)+3+{\sqrt {29}}}}\right|+C={\frac {1}{\sqrt {29}}}\ln \left|{\frac {-10x-6+3-{\sqrt {29}}}{-10x-6+3+{\sqrt {29}}}}\right|+C=$ $={\frac {1}{\sqrt {29}}}\ln \left|{\frac {-10x-3-{\sqrt {29}}}{-10x-3+{\sqrt {29}}}}\right|+C.$

Patikrinsime ar įstačius reikšmę x=0.1, atsakymai funkcijos ${\frac {1}{\sqrt {1-6x-9x^{2}}}}$ ir $({\frac {1}{\sqrt {29}}}\ln \left|{\frac {-10x-3-{\sqrt {29}}}{-10x-3+{\sqrt {29}}}}\right|+C)'$ sutaps:

${\frac {1}{\sqrt {1-6x-9x^{2}}}}={\frac {1}{\sqrt {1-0.1\cdot 6-9\cdot 0.1^{2}}}}={\frac {1}{\sqrt {1-0.6-0.09}}}={\frac {1}{\sqrt {0.31}}}={\frac {1}{0.556776436}}=1.796053021.$

$({\frac {1}{\sqrt {29}}}\ln \left|{\frac {-10x-3-{\sqrt {29}}}{-10x-3+{\sqrt {29}}}}\right|+C)'=({\frac {1}{\sqrt {29}}}(\ln \left|-10x-3-{\sqrt {29}}\right|-\ln \left|-10x-3+{\sqrt {29}}\right|)+C)'=$

$={\frac {1}{\sqrt {29}}}({\frac {-10}{-10x-3-{\sqrt {29}}}}-{\frac {-10}{-10x-3+{\sqrt {29}}}})={\frac {1}{\sqrt {29}}}({\frac {-10}{-10\cdot 0.1-3-{\sqrt {29}}}}-{\frac {-10}{-10\cdot 0.1-3+{\sqrt {29}}}})=$ $={\frac {1}{\sqrt {29}}}({\frac {-10}{-1-3-{\sqrt {29}}}}-{\frac {-10}{-1-3+{\sqrt {29}}}})={\frac {1}{5.385164807}}({\frac {-10}{-4-5.385164807}}+{\frac {10}{-4+5.385164807}})=$ $=0.185695338\cdot ({\frac {-10}{-9.385164807}}+{\frac {10}{1.385164807}})=0.185695338\cdot (1.06551139+7.219357545)=$ $=0.185695338\cdot 8.284868935=1.538461539.$

Jei trinaris butu virsuje integruotusi, butu daug paprasciau: $\int {\sqrt {1-6x-9x^{2}}}\;dx={\frac {1}{25}}\int (u^{2}+3u-5)du={\frac {1}{25}}[{u^{3} \over 3}+{3u^{2} \over 2}-5u+C={(-5x-3)^{3} \over 3}+{3(-5x-3)^{2} \over 2}-5(-5x-3)]+C=$ $={\frac {1}{25}}[{(-5x)^{3}-3\cdot (-5x)^{2}\cdot 3+3\cdot (-5x)\cdot 3^{2}-3^{3} \over 3}+{3[(-5x)^{2}-2\cdot (-5x)\cdot 3+3^{2}] \over 2}+25x+15]+C=$ $={\frac {1}{25}}[{-125x^{3}-9\cdot 25x^{2}+27\cdot (-5x)-27 \over 3}+{3[25x^{2}-6\cdot (-5x)+9] \over 2}+25x+15]+C=$ $={\frac {1}{25}}[{-125x^{3}-225x^{2}-135x-27 \over 3}+{3[25x^{2}+30x+9] \over 2}+25x+15]+C=$ $={\frac {1}{25}}[{-125x^{3} \over 3}-75x^{2}-45x-9+37.5x^{2}+45x+13.5+25x+15]+C={\frac {1}{25}}[{-125x^{3} \over 3}-37.5x^{2}+25x+19.5]+C.$

Patikrinsime ar istacius 0.1 reiksme i pradine funkcija ${\sqrt {1-6x-9x^{2}}}$ ir i funkcija ${\frac {1}{25}}({-125x^{3} \over 3}-37.5x^{2}+25x+19.5+C)'$ gausis tokie patys atsakymai.

${\frac {1}{25}}({-125x^{3} \over 3}-37.5x^{2}+25x+19.5+C)'={\frac {1}{25}}[-125x^{2}-75x+25]={\frac {1}{25}}[-125\cdot 0.1^{2}-75\cdot 0.1+25]=$

$={\frac {1}{25}}[-1.25-7.5+25]={\frac {1}{25}}\cdot 16.25=0.65.$ ${\sqrt {1-6x-9x^{2}}}={\sqrt {1-0.1\cdot 6-9\cdot 0.1^{2}}}={\sqrt {1-0.6-0.09}}={\sqrt {0.31}}=0.556776436.$

integruotos pradines funkcijos isvestines ir pradines funkcijos sulyginimas

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

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

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

$\left({\frac {1}{f}}\right)'={\frac {-f'}{f^{2}}},\qquad f\neq 0$

Palyginsime atsakymus integruotos funkcijos isvestines ir neintegruotos funkcijos, istacius abiais atvejais x=3.

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

$=2\cdot {\frac {(x+{\sqrt {x^{2}+x+1}})'}{x+{\sqrt {x^{2}+x+1}}}}-{\frac {3}{2}}\cdot {\frac {(1+2x+2{\sqrt {x^{2}+x+1}})'}{1+2x+2{\sqrt {x^{2}+x+1}}}}+{\frac {-3(2+4x+4{\sqrt {x^{2}+x+1}})'}{(2+4x+4{\sqrt {x^{2}+x+1}})^{2}}}+C'=$ $=2\cdot {\frac {1+{\frac {(x^{2}+x+1)'}{2{\sqrt {x^{2}+x+1}}}}}{x+{\sqrt {x^{2}+x+1}}}}-{\frac {3}{2}}\cdot {\frac {2+2\cdot {1 \over 2}\cdot {\frac {(x^{2}+x+1)'}{\sqrt {x^{2}+x+1}}}}{1+2x+2{\sqrt {x^{2}+x+1}}}}+{\frac {-3(4+4\cdot {1 \over 2}\cdot {\frac {(x^{2}+x+1)'}{\sqrt {x^{2}+x+1}}}}{(2+4x+4{\sqrt {x^{2}+x+1}})^{2}}}=$ $=2\cdot {\frac {1+{\frac {2x+1}{2{\sqrt {x^{2}+x+1}}}}}{x+{\sqrt {x^{2}+x+1}}}}-{\frac {3}{2}}\cdot {\frac {2+{\frac {2x+1}{\sqrt {x^{2}+x+1}}}}{1+2x+2{\sqrt {x^{2}+x+1}}}}+{\frac {-3(4+2\cdot {\frac {2x+1}{\sqrt {x^{2}+x+1}}})}{(2+4x+4{\sqrt {x^{2}+x+1}})^{2}}}=$ $=2\cdot {\frac {1+{\frac {2\cdot 3+1}{2{\sqrt {3^{2}+3+1}}}}}{3+{\sqrt {3^{2}+3+1}}}}-{\frac {3}{2}}\cdot {\frac {2+{\frac {2\cdot 3+1}{\sqrt {3^{2}+3+1}}}}{1+2\cdot 3+2{\sqrt {3^{2}+3+1}}}}+{\frac {-3(4+2\cdot {\frac {2\cdot 3+1}{\sqrt {3^{2}+3+1}}})}{(2+4\cdot 3+4{\sqrt {3^{2}+3+1}})^{2}}}=$ $=2\cdot {\frac {1+{\frac {7}{2{\sqrt {13}}}}}{3+{\sqrt {13}}}}-{\frac {3}{2}}\cdot {\frac {2+{\frac {7}{\sqrt {13}}}}{7+2{\sqrt {13}}}}+{\frac {-3(4+2\cdot {\frac {7}{\sqrt {13}}})}{(14+4{\sqrt {13}})^{2}}}={\frac {2+{\frac {7}{\sqrt {13}}}}{3+{\sqrt {13}}}}-{\frac {6+{\frac {21}{\sqrt {13}}}}{14+4{\sqrt {13}}}}-{\frac {12+{\frac {42}{\sqrt {13}}}}{(14+4{\sqrt {13}})^{2}}}=$ $={\frac {3.941450687}{6.605551275}}-{\frac {11.82435206}{28.4222051}}-{\frac {23.64870412}{807.8217428}}=0.596687622-0.416025147-0.029274656=0.151387818.$

${\frac {1}{x+{\sqrt {x^{2}+x+1}}}}={\frac {1}{3+{\sqrt {3^{2}+3+1}}}}={\frac {1}{3+{\sqrt {13}}}}={\frac {1}{6.605551275}}=0.151387818.$

integruotos pradines funkcijos isvestines ir pradines funkcijos sulyginimas 2

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

Patikriname ar atsakymai bus vienodi istacius x=4.

${\frac {1}{\sqrt {x^{2}+3}}}={\frac {1}{\sqrt {4^{2}+3}}}={\frac {1}{\sqrt {19}}}={\frac {1}{4.358898944}}=0.229415733.$

$(\ln |x+{\sqrt {x^{2}+3}}|+C)'={\frac {(x+{\sqrt {x^{2}+3}})'}{x+{\sqrt {x^{2}+3}}}}+C'={\frac {1+{1 \over 2}\cdot {\frac {(x^{2}+3)'}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {2x}{2{\sqrt {x^{2}+3}}}}}{x+{\sqrt {x^{2}+3}}}}={\frac {1+{\frac {x}{\sqrt {x^{2}+3}}}}{x+{\sqrt {x^{2}+3}}}}=$ $={\frac {1+{\frac {4}{\sqrt {4^{2}+3}}}}{4+{\sqrt {4^{2}+3}}}}={\frac {1+{\frac {4}{\sqrt {19}}}}{4+{\sqrt {19}}}}={\frac {1+{\frac {4}{4.358898944}}}{4+4.358898944}}={\frac {1+0.917662935}{8.358898944}}={\frac {1.917662935}{8.358898944}}=0.229415733.$

integruotos pradines funkcijos isvestines ir pradines funkcijos sulyginimas 3

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

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

Patikriname ar atsakymai bus vienodi istacius x=3.

$(-{\frac {1}{\sqrt {5}}}\ln |1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x)|+C)'=-{\frac {1}{\sqrt {5}}}\cdot {\frac {(1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x))'}{1-2{\sqrt {5}}({\sqrt {5x^{2}+x+1}}-{\sqrt {5}}x)}}=$

$=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {1}{2}}\cdot {\frac {2{\sqrt {5}}(10x+1)}{\sqrt {5x^{2}+x+1}}}+10}{1-{\sqrt {100x^{2}+20x+20}}+10x}}=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {{\sqrt {5}}(10\cdot 3+1)}{\sqrt {5\cdot 3^{2}+3+1}}}+10}{1-{\sqrt {100\cdot 3^{2}+20\cdot 3+20}}+10\cdot 3}}=$ $=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {31{\sqrt {5}}}{\sqrt {49}}}+10}{1-{\sqrt {900+80}}+30}}={\frac {1}{\sqrt {5}}}\cdot {\frac {{\frac {31{\sqrt {5}}}{7}}-10}{1-{\sqrt {980}}+30}}={\frac {1}{\sqrt {5}}}\cdot {\frac {-0,097413242}{-0.304951685}}=$ $={\frac {1}{\sqrt {5}}}\cdot 0,319438282=0,142857142.$

${\frac {1}{\sqrt {5x^{2}+x+1}}}={\frac {1}{\sqrt {5\cdot 3^{2}+3+1}}}={\frac {1}{\sqrt {5\cdot 9+4}}}={\frac {1}{\sqrt {49}}}={\frac {1}{7}}=0.142857142.$

O taip blogai:

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

$=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {1}{2}}\cdot {\frac {200x+20}{\sqrt {100x^{2}+20x+20}}}+10}{1-{\sqrt {100x^{2}+20x+20}}+10x}}=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {1}{2}}\cdot {\frac {200\cdot 3+20}{\sqrt {100\cdot 3^{2}+20\cdot 3+20}}}+10}{1-{\sqrt {100\cdot 3^{2}+20\cdot 3+20}}+10\cdot 3}}=$ $=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {1}{2}}\cdot {\frac {620}{\sqrt {920+80}}}+10}{1-{\sqrt {900+80}}+30}}=-{\frac {1}{\sqrt {5}}}\cdot {\frac {-{\frac {310}{\sqrt {1000}}}+10}{1-{\sqrt {980}}+30}}=-{\frac {1}{\sqrt {5}}}\cdot {\frac {0.196939253}{-0.304951685}}=$ $=-{\frac {1}{\sqrt {5}}}\cdot (-0.645804772)=0.288812674.$

integruotos pradines funkcijos isvestines ir pradines funkcijos sulyginimas 4

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

Patikriname ar atsakymai bus vienodi istacius $x={\frac {5}{2}}=2.5$.

$(-{\sqrt {\frac {2}{3}}}\arctan {\sqrt {\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}}}+C)'=-{\sqrt {\frac {2}{3}}}\cdot {\frac {{\sqrt {\frac {2}{3}}}\cdot (({\frac {3-x}{x-2}})^{1 \over 4})'}{1+({\sqrt {\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}}})^{2}}}=-{\frac {2}{3}}\cdot {\frac {{\frac {1}{4}}\cdot ({\frac {3-x}{x-2}})'}{({\frac {3-x}{x-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}})}}=$

$=-{\frac {1}{6}}\cdot {\frac {\frac {-x(x-2)-(3-x)x}{(x-2)^{2}}}{({\frac {3-x}{x-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}})}}=-{\frac {1}{6}}\cdot {\frac {\frac {-2.5(2.5-2)-(3-2.5)\cdot 2.5}{(2.5-2)^{2}}}{({\frac {3-2.5}{2.5-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-2.5}}}{3{\sqrt {2.5-2}}}})}}=-{\frac {1}{6}}\cdot {\frac {\frac {-2.5\cdot 0.5-0.5\cdot 2.5}{0.5^{2}}}{({\frac {0.5}{0.5}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {0.5}}}{3{\sqrt {0.5}}}})}}=$ $=-{\frac {1}{6}}\cdot {\frac {\frac {-2.5}{0.25}}{1^{3 \over 4}\cdot (1+{\frac {2}{3}})}}=-{\frac {1}{6}}\cdot {\frac {-10}{\frac {5}{3}}}=-{\frac {1}{6}}\cdot {\frac {-30}{5}}={\frac {1}{6}}\cdot 6=1.$

${\frac {1}{x{\sqrt {-x^{2}+5x-6}}}}={\frac {1}{2.5{\sqrt {-2.5^{2}+5\cdot 2.5-6}}}}={\frac {1}{2.5{\sqrt {-6.25+12.5-6}}}}=$

$={\frac {1}{2.5{\sqrt {-12.25+12.5}}}}={\frac {1}{2.5{\sqrt {0.25}}}}={\frac {1}{2.5\cdot 0.5}}={\frac {1}{1.25}}=0.8.$

Patikriname ar atsakymai bus vienodi istacius $x=2.4$.

$(-{\sqrt {\frac {2}{3}}}\arctan {\sqrt {\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}}}+C)'=-{\sqrt {\frac {2}{3}}}\cdot {\frac {{\sqrt {\frac {2}{3}}}\cdot (({\frac {3-x}{x-2}})^{1 \over 4})'}{1+({\sqrt {\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}}})^{2}}}=-{\frac {2}{3}}\cdot {\frac {{\frac {1}{4}}\cdot ({\frac {3-x}{x-2}})'}{({\frac {3-x}{x-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}})}}=$

$=-{\frac {1}{6}}\cdot {\frac {\frac {-x(x-2)-(3-x)x}{(x-2)^{2}}}{({\frac {3-x}{x-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-x}}}{3{\sqrt {x-2}}}})}}=-{\frac {1}{6}}\cdot {\frac {\frac {-2.4(2.4-2)-(3-2.4)\cdot 2.4}{(2.4-2)^{2}}}{({\frac {3-2.4}{2.4-2}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {3-2.4}}}{3{\sqrt {2.4-2}}}})}}=-{\frac {1}{6}}\cdot {\frac {\frac {-2.4\cdot 0.4-0.6\cdot 2.4}{0.4^{2}}}{({\frac {0.6}{0.4}})^{3 \over 4}\cdot (1+{\frac {2{\sqrt {0.6}}}{3{\sqrt {0.4}}}})}}=$ $=-{\frac {1}{6}}\cdot {\frac {\frac {-0.96-1.44}{0.16}}{1.5^{3 \over 4}\cdot (1+{\frac {2}{3}}\cdot 1.5)}}=-{\frac {1}{6}}\cdot {\frac {\frac {-2.4}{0.16}}{3.375^{1 \over 4}\cdot (1+1)}}=-{\frac {1}{6}}\cdot {\frac {-15}{{\sqrt {1.837117307}}\cdot 2}}={\frac {1}{12}}\cdot {\frac {15}{1.355403005}}=$ $={\frac {1}{12}}\cdot 11.0668192=0.922234933.$

${\frac {1}{x{\sqrt {-x^{2}+5x-6}}}}={\frac {1}{2.4{\sqrt {-2.4^{2}+5\cdot 2.4-6}}}}={\frac {1}{2.4{\sqrt {-5.76+12-6}}}}=$

$={\frac {1}{2.4{\sqrt {-11.76+12}}}}={\frac {1}{2.4{\sqrt {0.24}}}}={\frac {1}{2.4\cdot 0.489897948}}={\frac {1}{1.175755077}}=0.850517271.$

Update 1. Neteisingai integruojama buvo į arktangentą. Teisingai taip:

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

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

integruotos pradines funkcijos isvestines ir pradines funkcijos sulyginimas 5

$\int {\frac {x\;dx}{\sqrt {(7x-10-x^{2})^{3}}}}=\int {\frac {{\frac {5+2t^{2}}{1+t^{2}}}{\frac {-6t}{(1+t^{2})^{2}}}dt}{({\frac {3t}{1+t^{2}}})^{3}}}=\int {\frac {\frac {-6t(5+2t^{2})}{(1+t^{2})^{3}}}{({\frac {3t}{1+t^{2}}})^{3}}}dt=\int {\frac {-6t(5+2t^{2})}{27t^{3}}}dt=\int {\frac {-6(5+2t^{2})}{27t^{2}}}dt=$ $=\int (-{\frac {30}{27t^{2}}}-{\frac {12t^{2}}{27t^{2}}})dt={\frac {1}{27}}\int (-{\frac {30}{t^{2}}}-12)dt={\frac {1}{27}}({\frac {30}{t}}-12t)+C={\frac {1}{27}}({\frac {30}{\sqrt {\frac {(5-x)}{(x-2)}}}}-12{\sqrt {\frac {(5-x)}{(x-2)}}})+C=$ $={\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {(5-x)}{(x-2)}}}+C.$

Patikriname ar atsakymai bus vienodi istacius $x=3$.

$({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {1}{2}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot ({\frac {x-2}{5-x}})'-{\frac {12}{27}}\cdot {\frac {1}{2}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot ({\frac {5-x}{x-2}})'=$

$={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {x(5-x)-(x-2)\cdot (-x)}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x(x-2)-(5-x)x}{(x-2)^{2}}}=$ $={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {5x-x^{2}+x^{2}-2x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-x^{2}+2x-5x+x^{2}}{(x-2)^{2}}}=$ $={\frac {15}{27}}\cdot {\sqrt {\frac {5-x}{x-2}}}\cdot {\frac {3x}{(5-x)^{2}}}-{\frac {6}{27}}\cdot {\frac {\sqrt {x-2}}{\sqrt {5-x}}}\cdot {\frac {-3x}{(x-2)^{2}}}={\frac {1}{27}}({\sqrt {\frac {5-3}{3-2}}}\cdot {\frac {15\cdot 3\cdot 3}{(5-3)^{2}}}+{\frac {\sqrt {3-2}}{\sqrt {5-3}}}\cdot {\frac {18\cdot 3}{(3-2)^{2}}})=$ $={\frac {1}{27}}({\sqrt {\frac {2}{1}}}\cdot {\frac {135}{2^{2}}}+{\frac {\sqrt {1}}{\sqrt {2}}}\cdot {\frac {54}{1^{2}}})={\frac {1}{27}}({\frac {135{\sqrt {2}}}{4}}+{\frac {54}{\sqrt {2}}})={\frac {1}{27}}(47.72970773+38.18376618)=3.181980515.$

${\frac {x}{\sqrt {(7x-10-x^{2})^{3}}}}={\frac {3}{\sqrt {(7\cdot 3-10-3^{2})^{3}}}}={\frac {3}{\sqrt {(21-10-9)^{3}}}}={\frac {3}{\sqrt {2^{3}}}}={\frac {3}{\sqrt {8}}}=1.060660172.$

Pabandome gauti teisinga atsakyma isvestine darant kitokiu budu (ir istacius x=3), nors sis budas ligtais 95%, kad neteisingas:

$({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}-{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}}{({\sqrt {5-x}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {x-2}}{2{\sqrt {5-x}}}}-{\frac {\sqrt {5-x}}{2{\sqrt {x-2}}}}}{({\sqrt {x-2}})^{2}}}=$

$={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}-{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}}{({\sqrt {5-3}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {3-2}}{2{\sqrt {5-3}}}}-{\frac {\sqrt {5-3}}{2{\sqrt {3-2}}}}}{({\sqrt {3-2}})^{2}}}={\frac {30}{27}}\cdot {\frac {{\frac {\sqrt {2}}{2{\sqrt {1}}}}-{\frac {\sqrt {1}}{2{\sqrt {2}}}}}{({\sqrt {2}})^{2}}}-{\frac {12}{27}}\cdot {\frac {{\frac {\sqrt {1}}{2{\sqrt {2}}}}-{\frac {\sqrt {2}}{2{\sqrt {1}}}}}{({\sqrt {1}})^{2}}}=$ $={\frac {30}{27}}\cdot {\frac {\frac {2-1}{2{\sqrt {2}}}}{2}}-{\frac {12}{27}}\cdot {\frac {1-2}{2{\sqrt {2}}}}={\frac {30}{27}}\cdot {\frac {1}{4{\sqrt {2}}}}+{\frac {12}{27}}\cdot {\frac {1}{2{\sqrt {2}}}}={\frac {15}{54{\sqrt {2}}}}+{\frac {6}{27{\sqrt {2}}}}={\frac {15-12}{54{\sqrt {2}}}}={\frac {3}{54{\sqrt {2}}}}={\frac {1}{18{\sqrt {2}}}}=0.03928371.$

Pabandome gauti teisinga atsakyma isvestine darant kitokiu budu (ir istacius x=3), nors sis budas ligtais 95%, kad neteisingas:

$({\frac {30{\sqrt {x-2}}}{27{\sqrt {5-x}}}}-{\frac {12}{27}}{\sqrt {\frac {5-x}{x-2}}}+C)'=$

$={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {(x-2)'}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot ({\sqrt {5-x}})'}{({\sqrt {5-x}})^{2}}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {(5-x)'}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot ({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}})=$ $={\frac {30}{27}}({\frac {1}{2}}\cdot {\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot {\frac {1}{2}}\cdot (5-x)'}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {12}{27}}({\frac {1}{2}}\cdot {\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot {\frac {1}{2}}(x-2)'}{{\sqrt {x-2}}(x-2)}})=$ $={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {-{\sqrt {x-2}}\cdot (-1)}{{\sqrt {5-x}}\cdot (5-x)}})-{\frac {6}{27}}({\frac {-1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {-{\sqrt {5-x}}\cdot 1}{{\sqrt {x-2}}(x-2)}})=$ $={\frac {15}{27}}({\frac {1}{{\sqrt {x-2}}{\sqrt {5-x}}}}+{\frac {\sqrt {x-2}}{{\sqrt {5-x}}\cdot (5-x)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-x}}{\sqrt {x-2}}}}+{\frac {\sqrt {5-x}}{{\sqrt {x-2}}(x-2)}})=$ $={\frac {15}{27}}({\frac {1}{{\sqrt {3-2}}{\sqrt {5-3}}}}+{\frac {\sqrt {3-2}}{{\sqrt {5-3}}\cdot (5-3)}})+{\frac {6}{27}}({\frac {1}{{\sqrt {5-3}}{\sqrt {3-2}}}}+{\frac {\sqrt {5-3}}{{\sqrt {3-2}}(3-2)}})=$ $={\frac {15}{27}}({\frac {1}{\sqrt {2}}}+{\frac {1}{{\sqrt {2}}\cdot 2}})+{\frac {6}{27}}({\frac {1}{\sqrt {2}}}+{\frac {\sqrt {2}}{1}})={\frac {15}{27}}\cdot {\frac {2+1}{2{\sqrt {2}}}}+{\frac {6}{27}}\cdot {\frac {1+2}{\sqrt {2}}}={\frac {45}{54{\sqrt {2}}}}+{\frac {18}{27{\sqrt {2}}}}={\frac {5}{6{\sqrt {2}}}}+{\frac {2}{3{\sqrt {2}}}}=$

$=0.589255651+0.47140452=1.060660172.$

Sis budas pasirode esas teisingas. Cia buvo pasinaudota dviejomis formulemis:

$\left({fg}\right)'=f'g+fg';$

$\left({\frac {1}{f}}\right)'={\frac {-f'}{f^{2}}},\qquad f\neq 0.$

Diferencijavimas vyko tokiu budu, kad pavyzdiui, sioje funkcijoje ${\sqrt {\frac {5-x}{x-2}}}$ yra dvi funkcijos $f(x)=f={\sqrt {5-x}}$, o $g(x)=g={\frac {1}{\sqrt {x-2}}}$. Tuomet $(f(x))'=f'={\frac {1}{2{\sqrt {5-x}}}}$, o $(g(x))'=g'={\frac {-({\sqrt {x-2}})'}{({\sqrt {x-2}})^{2}}}={\frac {-1}{2{\sqrt {x-2}}({\sqrt {x-2}})^{2}}}.$

Neteisingai rastas dx pavyzdyje apie diferencialinių binomų integravimą

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

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

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

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

$=-{\frac {1}{2}}\int {\frac {dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-{\frac {1}{2}}\int {\frac {dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=$

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

$=-{1 \over 4}(\ln(t^{2}-1)-2\ln t).$ Kur $t={\sqrt {x^{2}+3 \over x^{2}}}.$

Iš tikro

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

Vis tiek su klaida

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

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

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

$=-\int {\frac {t\;dt}{\sqrt {(t^{4}-2t^{2}+1)+(t^{6}-3(t^{2})^{2}+3t^{2}-1^{3})}}}=-\int {\frac {t\;dt}{\sqrt {t^{6}-2t^{4}+t^{2}}}}=$

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

toliau integruojama kaip racionali funkcija.

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

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

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

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

Klaida čia:

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

Paraboloid (aptarimas) 19:55, 31 gruodžio 2022 (UTC)

Pasirodo, kad teisingai skaičiuojant gaunamas toks pat integralas $-\int {\frac {dt}{t^{2}-1}}.$ Tik dabar niekaip negaliu suprasti kaip buvo priskaičiuota teisingai, šiuo iš pažiūros klaidingu budu, kuris yra kažkaip teisingas. Arba ten tik galutinis atsakymas ($-\int {\frac {dt}{t^{2}-1}}$) teisingas, o ne pats skaičiavimo būdas...

Dabar atrodo supratau, kaip ten skaičiuota (ir ten teisingai skaičiuota)...