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Mariusz M

Rejestracja: 11 Sep 2010
Offline Ostatnio: May 07 2018 01:39

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#129813 cAŁAK 7

Napisane przez Mariusz M w 26.11.2017 - 03:43

Metoda Ostrogradskiego też będzie wymagała ośmiu współczynników , (po zastosowaniu wystarczy pobawić się licznikiem aby uzyskać dalszy rozkład)

Tutaj można pobawić się częściami aby uprościć sobie całkę

$8(4x^2-2x-3)=(8x-2)(4x-1)-26\\\\(8x-2)(4x-1)-8(4x^2-2x-3)=26\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^4}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^4}}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^4}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(\int{(4x-1)\frac{(8x-2)}{(4x^2-2x-3)^4}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{3}\frac{4x-1}{(4x^2-2x-3)^3}+\frac{4}{3}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{3}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{20}{3}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^3}\mbox{d}x}\\\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^3}}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(\int{(4x-1)\frac{(8x-2)}{(4x^2-2x-3)^3}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{2}\frac{4x-1}{(4x^2-2x-3)^2}+2\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=\frac{1}{26}\left(-\frac{1}{2}\frac{4x-1}{(4x^2-2x-3)^2}-6\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\ \\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\int{\frac{(8x-2)(4x-1)-8(4x^2-2x-3)}{(4x^2-2x-3)^2}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(\int{\frac{(8x-2)(4x-1)}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{(4x^2-2x-3)}{(4x^2-2x-3)^2}\mbox{d}x}\right)\\\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left((4x-1)\int{\frac{(8x-2)}{(4x^2-2x-3)^2}\mbox{d}x}-8\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(-\frac{4x-1}{4x^2-2x-3}+4\int{\frac{1}{4x^2-2x-3}\mbox{d}x}-8\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=\frac{1}{26}\left(-\frac{4x-1}{4x^2-2x-3}-4\int{\frac{1}{4x^2-2x-3}\mbox{d}x}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\int{\frac{4(4x-1+\sqrt{13})-4(4x-1-\sqrt{13})}{16x^2-8x-12}\mbox{d}x}\\ \\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\left(\int{\frac{4}{(4x-1-\sqrt{13})}\mbox{d}x}-\int{\frac{4}{(4x-1+\sqrt{13})}}\right)\\ \\\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}=-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\int{\frac{1}{(4x^2-2x-3)^2}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{3}{13}\left(-\frac{1}{26}\frac{4x-1}{4x^2-2x-3}-\frac{\sqrt{13}}{169}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}=-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}+\frac{3}{338}\frac{4x-1}{4x^2-2x-3}+\frac{3\sqrt{13}}{2197}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\int{\frac{1}{(4x^2-2x-3)^3}\mbox{d}x}\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}-\frac{10}{39}\left(-\frac{1}{52}\frac{4x-1}{(4x^2-2x-3)^2}+\frac{3}{338}\frac{4x-1}{4x^2-2x-3}+\frac{3\sqrt{13}}{2197}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}\right)\\\\\int{\frac{1}{(4x^2-2x-3)^4}\mbox{d}x}=-\frac{1}{78}\frac{4x-1}{(4x^2-2x-3)^3}+\frac{5}{1014}\frac{4x-1}{(4x^2-2x-3)^2}-\frac{5}{2197}\frac{4x-1}{4x^2-2x-3}-\frac{10\sqrt{13}}{28561}\ln{\left|\frac{4x-1-\sqrt{13}}{4x-1+\sqrt{13}}\right|}+C\\ \\$

#129812 Całka 30

Napisane przez Mariusz M w 26.11.2017 - 01:15

$\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=\frac{a_{1}x+a_{0}}{x^2+2x+2}+\int{\frac{b_{1}x+b_{0}}{x^2+2x+2}\mbox{d}x}\\\\\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}(x^2+2x+2)-(a_{1}x+a_{0})(2x+2)}{(x^2+2x+2)^2}+\frac{b_{1}x+b_{0}}{x^2+2x+2}\\\\\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})}{(x^2+2x+2)^2}+\frac{b_{1}x+b_{0}}{x^2+2x+2}\\\\\\\frac{x}{\left(x^2+2x+2\right)^2}=\frac{a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})+(b_{1}x+b_{0})(x^2+2x+2)}{(x^2+2x+2)^2}\\\\x=a_{1}x^2+2a_{1}x+2a_{1}-(2a_{1}x^2+2a_{1}x+2a_{0}x+2a_{0})+(b_{1}x+b_{0})(x^2+2x+2)\\\\x=a_{1}x^2+2a_{1}x+2a_{1}-2a_{1}x^2-2a_{1}x-2a_{0}x-2a_{0}+b_{1}x^3+2b_{1}x^2+2b_{1}x+b_{0}x^2+2b_{0}x+2b_{0}\\\\x=b_{1}x^3+(2b_{1}+b_{0}-a_{1})x^2+(2b_{1}+2b_{0}-2a_{0})x+2b_{0}+2a_{1}-2a_{0}\\\\\begin{cases}b_{1}=0\\2b_{1}+b_{0}-a_{1}=0\\2b_{1}+2b_{0}-2a_{0}=1\\2b_{0}+2a_{1}-2a_{0}=0 \end{cases}\\ \\\begin{cases}b_{1}=0\\b_{0}=a_{1}\\2a_{1}-2a_{0}=1\\4a_{1}-2a_{0}=0 \end{cases}\\ \\\begin{cases}b_{1}=0\\b_{0}=a_{1}\\2a_{1}=-1\\a_{0}=2a_{1} \end{cases}\\ \\\begin{cases}b_{1}=0\\2b_{0}=-1\\2a_{1}=-1\\a_{0}=-1 \end{cases}\\ \\\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=-\frac{1}{2}\frac{x+2}{x^2+2x+2}-\frac{1}{2}\int{\frac{\mbox{d}x}{(x+1)^2+1}}\\\\\int{\frac{x}{\left(x^2+2x+2\right)^2}\mbox{d}x}=-\frac{1}{2}\frac{x+2}{x^2+2x+2}-\frac{1}{2}\arctan{(x+1)}+C\\ \\$

Można jeszcze bawić się całkowaniem przez części

#129608 Całka funkcji trygonometrycznej

Napisane przez Mariusz M w 18.09.2017 - 23:26

$\int{\sqrt{\tan^{2}{\left(x\right)}+4}\mbox{d}x}\\\\\sqrt{\tan^{2}{\left(x\right)}+4}=t-\tan{\left(x\right)}\\\\\tan^{2}{\left(x\right)}+4=t^2-2t\tan{\left(x\right)}+\tan^{2}{\left(x\right)}\\\\4=t^2-2t\tan{\left(x\right)}\\\\2t\tan{\left(x\right)}=t^2-4\\\\\tan{\left(x\right)}=\frac{t^2-4}{2t}\\\\\left(1+\tan^2{\left(x\right)}\right)\mbox{d}x=\frac{2t\cdot 2t-2\left(t^2-4\right)}{4t^2}\mbox{d}t\\\\\left(1+\frac{t^4-8t^2+16}{4t^2}\right)\mbox{d}x=\frac{t^2+4}{2t^2}\mbox{d}t\\\\\frac{t^4-4t^2+16}{4t^2}\mbox{d}x=\frac{t^2+4}{2t^2}\mbox{d}t\\\\\mbox{d}x=\frac{t^2+4}{2t^2}\cdot\frac{4t^2}{t^4-4t^2+16}\mbox{d}t\\\\\mbox{d}x=2\frac{t^2+4}{t^4-4t^2+16}\mbox{d}t\\\\\sqrt{\tan^{2}{\left(x\right)}+4}=t-\frac{t^2-4}{2t}\\\\\sqrt{\tan^{2}{\left(x\right)}+4}=\frac{t^2+4}{2t}\\\\\int{\frac{\left(t^2+4\right)^2}{t\left(t^4-4t^2+16\right)}\mbox{d}t}\\\\\frac{1}{2}\int{\frac{2t\left(t^2+4\right)^2}{t^2\left(t^4-4t^2+16\right)}\mbox{d}t}\\\\u=t^2\\\\\mbox{d}u=2t\mbox{d}t\\\\\frac{1}{2}\int{\frac{\left(u+4\right)^2}{u\left(u^2-4u+16\right)}\mbox{d}u}\\\\\frac{1}{2}\int{\frac{u^2+8u+16}{u\left(u^2-4u+16\right)}\mbox{d}u}\\\\\frac{1}{2}\int{\frac{u^2-4u+16+12u}{u\left(u^2-4u+16\right)}\mbox{d}u}\\\\\frac{1}{2}\left(\int{\frac{\mbox{d}u}{u}}+\int{\frac{12}{u^2-4u+16}\mbox{d}u}\right)\\\\\frac{1}{2}\left(\int{\frac{\mbox{d}u}{u}}+\int{\frac{12}{\left(u-2\right)^2+12}\mbox{d}u}\right)\\\\\int{\frac{12}{\left(u-2\right)^2+12}\mbox{d}u}\\\\u-2=2\sqrt{3}v\\\\\mbox{d}u=2\sqrt{3}\mbox{d}v\\\\24\sqrt{3}\int{\frac{\mbox{d}v}{12v^2+12}}\\\\2\sqrt{3}\int{\frac{\mbox{d}v}{v^2+1}}\\\\\frac{1}{2}\left(\ln{\left|u\right|}+2\sqrt{3}\arctan{\left(\frac{u-2}{2\sqrt{3}}\right)}\right)\\\\\ln{\left|t\right|}+\sqrt{3}\arctan{\left(\frac{t^2-2}{2\sqrt{3}}\right)}\\\\\ln{\left|\tan{\left(x\right)}+\sqrt{\tan^{2}{\left(x\right)}+4}\right|}+\sqrt{3}\arctan{\left(\frac{1+\tan^{2}{\left(x\right)}+\tan{\left(x\right)}\sqrt{\tan^{2}{\left(x\right)}+4}}{\sqrt{3}}\right)}+C\\\\$

Można było jednym podstawieniem, te pozostałe podstawienia są tylko dla wygody

#129232 Równanie różniczkowe

Napisane przez Mariusz M w 21.04.2017 - 12:16

Można też skorzystać z podanej postaci czynnika całkującego

$\mu=\mu\left[\omega\left(x,y\right)\right]\\ \\P=P\left(x,y\right)\\ \\Q=Q\left(x,y\right)\\ \\\frac{\partial \mu P}{\partial y}=\frac{\partial \mu Q}{\partial x}\\ \\\frac{\partial \mu}{\partial y}P+\mu\frac{\partial P}{\partial y}=\frac{\partial \mu}{\partial x}Q+\mu\frac{\partial Q}{\partial x}\\ \\\frac{\partial \mu}{\partial y}P-\frac{\partial \mu}{\partial x}Q=\mu\frac{\partial Q}{\partial x}-\mu\frac{\partial P}{\partial y}\\ \\\frac{\partial \mu}{\partial y}P-\frac{\partial \mu}{\partial x}Q=\mu\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ \\\frac{\mbox{d}\mu}{\mbox{d}\omega}\cdot\frac{\partial \omega}{\partial y}P-\frac{\mbox{d}\mu}{\mbox{d}\omega}\cdot\frac{\partial \omega}{\partial x}Q=\mu\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ \\\frac{\mbox{d}\mu}{\mbox{d}\omega}\left(\frac{\partial \omega}{\partial y}P-\frac{\partial \omega}{\partial x}Q\right)=\mu\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\\ \\\frac{\mbox{d}\mu}{\mu}=\frac{\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}}{\frac{\partial \omega}{\partial y}P-\frac{\partial \omega}{\partial x}Q}\mbox{d}\omega\\ \\\frac{\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}}{\frac{\partial \omega}{\partial y}P-\frac{\partial \omega}{\partial x}Q}=\varphi\left(\omega\right)\\ \\\frac{\mbox{d}\mu}{\mu}=\varphi\left(\omega\right)\mbox{d}\omega \\$

#129199 Całaka 14

Napisane przez Mariusz M w 04.04.2017 - 22:53

$\int{\sqrt{1-x^2}\arcsin{x}\mbox{d}x}=\int{\frac{1-x^2}{\sqrt{1-x^2}}\arcsin{x}\mbox{d}x}\\\\=\int{\frac{\arcsin{x}}{\sqrt{1-x^2}}\mbox{d}x}+\int{\frac{-x}{\sqrt{1-x^2}}\left(x\arcsin{x}\right)\mbox{d}x}\\\\=\int{\frac{\arcsin{x}}{\sqrt{1-x^2}}\mbox{d}x}+x\sqrt{1-x^2}\arcsin{x}-\int{\sqrt{1-x^2}\left(\arcsin{x}+\frac{x}{\sqrt{1-x^2}}\right)\mbox{d}x}\\\\\int{\sqrt{1-x^2}\arcsin{x}\mbox{d}x}=\int{\frac{\arcsin{x}}{\sqrt{1-x^2}}\mbox{d}x}+x\sqrt{1-x^2}\arcsin{x}-\int{\sqrt{1-x^2}\arcsin{x}\mbox{d}x}-\int{x\mbox{d}x}\\\\2\int{\sqrt{1-x^2}\arcsin{x}\mbox{d}x}=\frac{1}{2}\arcsin^2{x}+x\sqrt{1-x^2}\arcsin{x}-\frac{1}{2}x^2+C\\\\\int{\sqrt{1-x^2}\arcsin{x}\mbox{d}x}=\frac{1}{4}\arcsin^2{x}+\frac{1}{2}x\sqrt{1-x^2}\arcsin{x}-\frac{1}{4}x^2+C\\\\$

#129198 Całka 18

Napisane przez Mariusz M w 04.04.2017 - 20:03

Dosyć dobry efekt daje pierwsze podstawienie Eulera ale jak to modnie jest unikać tych podstawień

$\int{\frac{5x^2-2x+10}{\sqrt{3x^2-5x+8}}\mbox{d}x}\\\\\sqrt{3}\int{\frac{5x^2-2x+10}{\sqrt{9x^2-15x+24}}\mbox{d}x}\\\\\sqrt{9x^2-15x+24}=t-3x\\\\9x^2-15x+24=t^2-6tx+9x^2\\-15x+24=t^2-6tx\\\\6tx-15x=t^2-24\\\\x\left(6t-15\right)=t^2-24\\\\x=\frac{t^2-24}{6t-15}\\\\t-3x=\frac{6t^2-15t-3t^2+72}{6t-15}=\frac{3t^2-15t+72}{6t-15}\\\\\mbox{d}x=\frac{2t\left(6t-15\right)-6\left(t^2-24\right)}{\left(6t-15\right)^2}\mbox{d}t\\\\\mbox{d}x=\frac{6t^2-30t+144}{\left(6t-15\right)^2}\mbox{d}t\\\\\sqrt{3}\int{\left(5\frac{\left(t^2-24\right)^2}{\left(6t-15\right)^2}-2\frac{t^2-24}{6t-15}+10\right)\frac{6t-15}{3t^2-15t+72}\cdot\frac{2\left(3t^2-15t+72\right)}{\left(6t-15\right)^2}\mbox{d}t}\\\\\sqrt{3}\int{\left(\frac{5\left(t^4-48t^2+576\right)-2\left(t^2-24\right)\left(6t-15\right)+10\left(36t^2-180t+225\right)}{\left(6t-15\right)^2}\right)\cdot\frac{2}{6t-15}\mbox{d}t}\\\\2\sqrt{3}\int{\frac{5t^4-240t^2+2880-2\left(6t^3-15t^2-144t+360\right)+360t^2-1800t+2250}{\left(6t-15\right)^3}\mbox{d}t}\\\\2\sqrt{3}\int{\frac{5t^4-240t^2+2880-12t^3+30t^2+288t-720+360t^2-1800t+2250}{\left(6t-15\right)^3}\mbox{d}t}\\\\2\sqrt{3}\int{\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(6t-15\right)^3}\mbox{d}t}\\\\\frac{2\sqrt{3}}{27}\int{\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^3}\mbox{d}t}\\\\-\frac{\sqrt{3}}{54}\int{\left(5t^4-12t^3+150t^2-1512t+4410\right)\frac{\left(-4\right)}{\left(2t-5\right)^3}\mbox{d}t}\\\\-\frac{\sqrt{3}}{54}\left(\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^2}-\int{\frac{20t^3-36t^2+300t-1512}{\left(2t-5\right)^2}\mbox{d}t}\right)\\\\-\frac{\sqrt{3}}{54}\left(\frac{5t^4-12t^3+150t^2-1512t+4410}{\left(2t-5\right)^2}+\int{\left(10t^3-18t^2+150t-756\right)\frac{\left(-2\right)}{\left(2t-5\right)^2}\mbox{d}t}\right)\\\\-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-\int{\frac{60t^2-72t+300}{2t-5}\mbox{d}t}\right)\\\\-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-\int{\left(30t+39\right)\mbox{d}t}-\int{\frac{495}{2t-5}\mbox{d}t}\right)\\\\-\frac{\sqrt{3}}{108}\left(\frac{10t^4-24t^3+300t^2-3024t+8820}{\left(2t-5\right)^2}+\frac{20t^3-36t^2+300t-1512}{2t-5}-15t^2-39t-\frac{495}{2}\ln{\left|2t-5\right|}\right)+C\\\\$

#129197 Całka 23, Ostrogradskiego można

Napisane przez Mariusz M w 04.04.2017 - 19:22

Jeśli chodzi o zalety metody Ostrogradskiego to nie wymaga ona rozkładu mianownika na czynniki

co pozwoli zredukować jego stopień

Aby uzyskać mianowniki wystarczy policzyć $NWD\left(M(x),M'(x)\right)$

algorytmem kolejnych dzieleń

Jeżeli mamy podany rozkład mianownika na czynniki to zalety stosowania metody Ostrogradskiego ujawniają się gdy

mianownik ma wielokrotne pierwiastki zespolone

#129196 Całka pierwiastek

Napisane przez Mariusz M w 04.04.2017 - 12:59

$\int{x^{\frac{1}{3}}\left(7-3x^2\right)^{\frac{1}{3}}\mbox{d}x}\\\\m=\frac{1}{3}\\\\n=2\\\\p=\frac{1}{3}\\\\\frac{m+1}{n}=\frac{2}{3}\\\\\frac{m+1}{n}+p=1 \in \mathbb{Z}\\\\t^3=\frac{7-3x^2}{x^2}\\$

#129117 Rozwiąż równanie różnicowe

Napisane przez Mariusz M w 19.03.2017 - 04:58

$y_{n+2}-4y_{n+1}+4y_{n}=0 \qquad y_{0}=3\\y_{1}=2\\\\y_{n}-4y_{n-1}+4y_{n-2}=0 \qquad y_{0}=3\\y_{1}=2\\\\y_{n}=4y_{n-1}-4y_{n-2}=0 \qquad y_{0}=3\\y_{1}=2\\\\Y\left(t\right)=\sum_{n=0}^{\infty}y_{n}t^n\\\\\sum_{n=2}^{\infty}y_{n}t^{n}=\sum_{n=2}^{\infty}4y_{n-1}t^n+\sum_{n=2}^{\infty}{-4y_{n-2}t^n}\\\\\sum_{n=2}^{\infty}y_{n}t^{n}=4t\sum_{n=2}^{\infty}y_{n-1}t^{n-1}-4t^2\sum_{n=2}^{\infty}{y_{n-2}t^{n-2}}\\\\\sum_{n=2}^{\infty}y_{n}t^{n}=4t\sum_{n=1}^{\infty}y_{n}t^{n}-4t^2\sum_{n=0}^{\infty}{y_{n}t^{n}}\\\\\sum_{n=0}^{\infty}y_{n}t^{n}-3-2t=4t\left(\sum_{n=0}^{\infty}y_{n}t^{n}-3\right)-4t^2\sum_{n=0}^{\infty}{y_{n}t^{n}}\\\\Y\left(t\right)-3-2t=4tY\left(t\right)-12t-4t^2Y\left(t\right)\\\\Y\left(t\right)\left(1-4t+4t^2\right)=-10t+3\\\\Y\left(t\right)=\frac{-10t+3}{\left(1-2t\right)^2}\\\\Y\left(t\right)=\frac{5-10t-2}{\left(1-2t\right)^2}\\\\Y\left(t\right)=\frac{5}{1-2t}-\frac{2}{\left(1-2t\right)^2}\\\\\sum_{n=0}^{\infty}2^{n}t^n=\frac{1}{1-2t}\\\\\frac{\mbox{d}}{\mbox{d}t}\left(\sum_{n=0}^{\infty}2^{n}t^n\right)=\frac{\mbox{d}}{\mbox{d}t}\left(\frac{1}{1-2t}\right)\\\\\sum_{n=0}^{\infty}n2^{n}t^{n-1}=-\frac{1}{\left(1-2t\right)^2}\cdot\left(-2\right)\\\\\sum_{n=1}^{\infty}n2^{n}t^{n-1}=\frac{2}{\left(1-2t\right)^2}\\\\\sum_{n=0}^{\infty}\left(n+1\right)2^{n+1}t^{n}=\frac{2}{\left(1-2t\right)^2}\\\\\sum_{n=0}^{\infty}\left(n+1\right)2^{n}t^{n}=\frac{1}{\left(1-2t\right)^2}\\\\Y\left(t\right)=\sum_{n=0}^{\infty}5\cdot 2^{n}t^{n}-2\sum_{n=0}^{\infty}\left(n+1\right)2^{n}t^n\\\\y_{n}=5\cdot 2^{n}-2\left(n+1\right)\cdot 2^{n}\\\\y_{n}=\left(3-2n\right)\cdot 2^{n}\\$

#129062 Całka 29

Napisane przez Mariusz M w 05.03.2017 - 06:11

Proponuję rozbić na dwie całki w jednej licznik skróci się z mianownikiem a drugą można będzie policzyć całkując przez części

Najpierw całkujemy przez części a później dodajemy całki

Przydatne będzie też podstawienie $x+\frac{1}{2}=\frac{3}{2}\sqrt{11}t$

Jak znalazłem ten rozkład ?

Podzieliłem licznik przez pochodną trójmianu kwadratowego z mianownika

Otrzymałem resztę która była stałą, do tej stałej dodałem i odjąłem taki składnik aby otrzymać postać kanoniczną trójmianu kwadratowego z mianownika

$\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=-\frac{1}{198}\int{\frac{x^2+x+25}{\left(x^2+x+25\right)^2}\mbox{d}x}+\frac{1}{198}\int{\left(99x^2-49x+25\right)\cdot\frac{2x+1}{\left(x^2+x+25\right)^2}\mbox{d}x}\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=-\frac{1}{198}\int{\frac{1}{\left(x^2+x+25\right)}\mbox{d}x}+\frac{1}{198}\int{\left(99x^2-49x+25\right)\cdot\frac{2x+1}{\left(x^2+x+25\right)^2}\mbox{d}x}\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=-\frac{1}{198}\int{\frac{1}{\left(x^2+x+25\right)}\mbox{d}x}-\frac{1}{198}\frac{99x^2-49x+25}{x^2+x+25}+\frac{1}{198}\int{\frac{198x-49}{x^2+x+25}\mbox{d}x}\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=-\frac{1}{198}\frac{99x^2-49x+25}{x^2+x+25}+\frac{1}{99}\int{\frac{99x-25}{x^2+x+25}\mbox{d}x}\\\\\frac{1}{99}\int{\frac{99x-25}{x^2+x+25}\mbox{d}x}=\frac{1}{99}\int{\frac{99x-25}{\left(x+\frac{1}{2}\right)^2+\frac{99}{4}}\mbox{d}x}\\\\x+\frac{1}{2}=\frac{3}{2}\sqrt{11}t\\\\\mbox{d}x=\frac{3}{2}\sqrt{11}\mbox{d}t\\\\\frac{\sqrt{11}}{66}\int{\frac{99\left(\frac{3}{2}\sqrt{11}t-\frac{1}{2}\right)-25}{\frac{99}{4}t^2+\frac{99}{4}}\mbox{d}t}\\\\\frac{\sqrt{11}}{66}\int{\frac{\frac{297}{2}\sqrt{11}t-\frac{149}{2}}{\frac{99}{4}t^2+\frac{99}{4}}\mbox{d}t}\\\\\frac{\sqrt{11}}{4 \cdot 33}\cdot\frac{4}{99}\int{\frac{297\sqrt{11}t-149}{t^2+1}\mbox{d}t}\\\\\frac{297\cdot 11}{3267}\int{\frac{t}{t^2+1}\mbox{d}t}-\frac{149\sqrt{11}}{3267}\int{\frac{1}{t^2+1}\mbox{d}t}\\\\=\int{\frac{t}{t^2+1}\mbox{d}t}-\frac{149\sqrt{11}}{3267}\int{\frac{1}{t^2+1}\mbox{d}t}\\\\=\frac{1}{2}\ln{\left|t^2+1\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(t\right)}+C\\\\\frac{1}{2}\ln{\left|x^2+x+25\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(\frac{\sqrt{11}}{33}\left(2x+1\right)\right)}+C\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=-\frac{1}{198}\frac{99x^2-49x+25}{x^2+x+25}+\frac{1}{2}\ln{\left|x^2+x+25\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(\frac{\sqrt{11}}{33}\left(2x+1\right)\right)}+C\\\\$

Wynik można jeszcze uprościć dzieląc licznik przez mianownik

Można też liczyć sposobem podanym przez Ostrogradskiego

$\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=\frac{a_{1}x+a_{0}}{x^2+x+25}+\int{\frac{b_{1}x+b_{0}}{x^2+x+25}\mbox{d}x}\\\\\frac{x^3}{\left(x^2+x+25\right)^2}=\frac{a_{1}\left(x^2+x+25\right)-\left(a_{1}x+a_{0}\right)\left(2x+1\right)}{\left(x^2+x+25\right)^2}+\frac{b_{1}x+b_{0}}{x^2+x+25}\\\\\frac{x^3}{\left(x^2+x+25\right)^2}=\frac{a_{1}\left(x^2+x+25\right)-\left(a_{1}x+a_{0}\right)\left(2x+1\right)+\left(b_{1}x+b_{0}\right)\left(x^2+x+25\right)}{\left(x^2+x+25\right)^2}\\\\x^3=a_{1}\left(x^2+x+25\right)-\left(a_{1}x+a_{0}\right)\left(2x+1\right)+\left(b_{1}x+b_{0}\right)\left(x^2+x+25\right)\\\\x^3=a_{1}x^2+a_{1}x+25a_{1}-2a_{1}x^2-a_{1}x-2a_{0}x-a_{0}+b_{1}x^3+b_{1}x^2+25b_{1}x+b_{0}x^2+b_{0}x+25b_{0}\\\\x^3=b_{1}x^3+\left(b_{1}+b_{0}-a_{1}\right)x^2+\left(25b_{1}+b_{0}-2a_{0}\right)x+25b_{0}+25a_{1}-a_{0}\\\\\begin{cases}b_{1}=1\\b_{0}=a_{1}-1\\24+a_{1}=2a_{0}\\100a_{1}-50-24-a_{1}=0\end{cases}\\\\\begin{cases}b_{1}=1\\\b_{0}=a_{1}-1\\24+a_{1}=2a_{0}\\99a_{1}=74\end{cases}\\\\\begin{cases}b_{1}=1\\99b_{0}=-25\\99a_{0}=1225\\99a_{1}=74\end{cases}\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=\frac{1}{99}\frac{74x+1225}{x^2+x+25}+\frac{1}{99}\int{\frac{99x-25}{x^2+x+25}\mbox{d}x}\\\\\frac{1}{99}\int{\frac{99x-25}{x^2+x+25}\mbox{d}x}=\frac{1}{99}\int{\frac{99x-25}{\left(x+\frac{1}{2}\right)^2+\frac{99}{4}}\mbox{d}x}\\\\x+\frac{1}{2}=\frac{3}{2}\sqrt{11}t\\\\\mbox{d}x=\frac{3}{2}\sqrt{11}\mbox{d}t\\\\\frac{\sqrt{11}}{66}\int{\frac{99\left(\frac{3}{2}\sqrt{11}t-\frac{1}{2}\right)-25}{\frac{99}{4}t^2+\frac{99}{4}}\mbox{d}t}\\\\\frac{\sqrt{11}}{66}\int{\frac{\frac{297}{2}\sqrt{11}t-\frac{149}{2}}{\frac{99}{4}t^2+\frac{99}{4}}\mbox{d}t}\\\\\frac{\sqrt{11}}{4 \cdot 33}\cdot\frac{4}{99}\int{\frac{297\sqrt{11}t-149}{t^2+1}\mbox{d}t}\\\\\frac{297\cdot 11}{3267}\int{\frac{t}{t^2+1}\mbox{d}t}-\frac{149\sqrt{11}}{3267}\int{\frac{1}{t^2+1}\mbox{d}t}\\\\=\int{\frac{t}{t^2+1}\mbox{d}t}-\frac{149\sqrt{11}}{3267}\int{\frac{1}{t^2+1}\mbox{d}t}\\\\=\frac{1}{2}\ln{\left|t^2+1\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(t\right)}+C\\\\\frac{1}{2}\ln{\left|x^2+x+25\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(\frac{\sqrt{11}}{33}\left(2x+1\right)\right)}+C\\\\\int{\frac{x^3}{\left(x^2+x+25\right)^2}\mbox{d}x}=\frac{1}{99}\frac{74x+1225}{x^2+x+25}+\frac{1}{2}\ln{\left|x^2+x+25\right|}-\frac{149\sqrt{11}}{3267}\arctan{\left(\frac{\sqrt{11}}{33}\left(2x+1\right)\right)}+C\\\\$

#129051 Całka 24

Napisane przez Mariusz M w 03.03.2017 - 08:31

$\int{\frac{\mbox{d}x}{\left(x^2+x+1\right)\sqrt{x^2+x-1}}}\\\\\sqrt{x^2+x-1}=t-x\\\\x^2+x-1=t^2-2tx+x^2\\\\x-1=t^2-2tx\\\\2tx+x=t^2+1\\\\x\left(2t+1\right)=t^2+1\\\\x=\frac{t^2+1}{2t+1}\\\\t-x=\frac{2t^2+t-t^2-1}{2t+1}=\frac{t^2+t-1}{2t+1}\\\\x^2+x+1=\frac{t^4+2t^2+1+2t^3+t^2+2t+1+4t^2+4t+1}{\left(2t+1\right)^2}\\ \\x^2+x+1=\frac{t^4+2t^3+7t^2+6t+3}{\left(2t+1\right)^2}\\\\\mbox{d}x=\frac{2t\left(2t+1\right)-2\left(t^2+1\right)}{\left(2t+1\right)^2}\mbox{d}t\\\\\mbox{d}x=\frac{2t^2+2t-2}{\left(2t+1\right)^2}\mbox{d}t\\\\\int{\frac{\left(2t+1\right)^2}{t^4+2t^3+7t^2+6t+3}\cdot\frac{2t+1}{t^2+t-1}\cdot\frac{2\left(t^2+t-1\right)}{\left(2t+1\right)^2}\mbox{d}t}\\\\\int{\frac{4t+2}{t^4+2t^3+7t^2+6t+3}\mbox{d}t}=\int{\frac{4t+2}{\left(t^2+t+3-\sqrt{6}\right)\left(t^2+t+3+\sqrt{6}\right)}\mbox{d}t}\\\\=\int{\frac{\sqrt{6}}{6}\left(\frac{2t+1}{t^2+t+3-\sqrt{6}}-\left(\frac{2t+1}{t^2+t+3-\sqrt{6}}\right)\right)\mbox{d}t}\\\\=\frac{\sqrt{6}}{6}\int{\frac{2t+1}{t^2+t+3-\sqrt{6}}\mbox{d}t}-\frac{\sqrt{6}}{6}\int{\frac{2t+1}{t^2+t+3+\sqrt{6}}\mbox{d}t}\\\\=\frac{\sqrt{6}}{6}\ln{\left|\frac{t^2+t+3-\sqrt{6}}{t^2+t+3+\sqrt{6}}\right|}+C\\$

#128996 całka 28

Napisane przez Mariusz M w 28.02.2017 - 05:35

$\int{\frac{\mbox{d}x}{\left(x^2-1\right)^2}e^{x}\left(1+2x-x^2\right)}\\ \\\int{\frac{1-x^2}{\left(x^2-1\right)^2}e^{x}\mbox{d}x}+\int{\frac{2x}{\left(x^2-1\right)^2}e^{x}\mbox{d}x}\\ \\=-\int{\frac{e^{x}}{x^2-1}\mbox{d}x}+\int{\frac{2x}{\left(x^2-1\right)^2}e^{x}\mbox{d}x}\\ \\=-\int{\frac{e^{x}}{x^2-1}\mbox{d}x}-\frac{e^{x}}{x^2-1}+\int{\frac{e^{x}}{x^2-1}\mbox{d}x}\\ \\=-\frac{e^{x}}{x^2-1}+C$

#128938 Całka wymierna, Ostrogradskiego

Napisane przez Mariusz M w 21.02.2017 - 22:02

To podstawienie jest kiepskie pod względem metodyki nauczania bo całki z funkcji wymiernej poznajemy wcześniej

ale za to jakie modne amerykańskie

$\int{\frac{3x+5}{\left(x^2+1\right)^3}\mbox{d}x}=\frac{a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}}{\left(x^2+1\right)^2}+\int{\frac{b_{1}x+b_{0}}{x^2+1}\mbox{d}x}\\\\\frac{3x+5}{\left(x^2+1\right)^3}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+1\right)^2-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\left(1+x^2\right)\cdot 4x}{\left(1+x^2\right)^4}+\frac{b_{1}x+b_{0}}{x^2+1}\\\\\frac{3x+5}{\left(x^2+1\right)^3}=\frac{\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+1\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\cdot 4x+\left(b_{1}x+b_{0}\right)\left(x^4+2x^2+1\right)}{\left(x^2+1\right)^3}\\\\3x+5=\left(3a_{3}x^2+2a_{2}x+a_{1}\right)\left(x^2+1\right)-\left(a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}\right)\cdot 4x+\left(b_{1}x+b_{0}\right)\left(x^4+2x^2+1\right)\\\\3x+5=3a_{3}x^4+2a_{2}x^3+a_{1}x^2+3a_{3}x^2+2a_{2}x+a_{1}-4a_{3}x^{4}-4a_{2}x^{3}-4a_{1}x^2-4a_{0}x+b_{1}x^5+2b_{1}x^3+b_{1}x+b_{0}x^4+2b_{0}x^2+b_{0}\\\\3x+5=b_{1}x^5+\left(b_{0}-a_{3}\right)x^4+\left(2b_{1}-2a_{2}\right)x^3+\left(2b_{0}+3a_{3}-3a_{1}\right)x^2+\left(b_{1}+2a_{2}-4a_{0}\right)x+a_{1}+b_{0}\\\\\begin{cases}b_{1}=0\\b_{0}-a_{3}=0\\2b_{1}-2a_{2}=0\\2b_{0}+3a_{3}-3a_{1}=0\\b_{1}+2a_{2}-4a_{0}=3\\a_{1}+b_{0}=5\end{cases}\\\\\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=0\\5a_{3}-3a_{1}=0\\-4a_{0}=3\\a_{3}+a_{1}=5\end{cases}\\\\\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=0\\5a_{3}-3a_{1}=0\\-4a_{0}=3\\3a_{3}+3a_{1}=15\end{cases}\\\\\begin{cases}b_{1}=0\\b_{0}=a_{3}\\a_{2}=0\\3a_{1}=5a_{3}\\-4a_{0}=3\\8a_{3}=15\end{cases}\\\\\begin{cases}b_{1}=0\\b_{0}=\frac{15}{8}\\a_{3}=\frac{15}{8}\\a_{2}=0\\a_{1}=\frac{25}{8}\\a_{0}=-\frac{6}{8}\end{cases}\\\\\int{\frac{3x+5}{\left(x^2+1\right)^3}\mbox{d}x}=\frac{1}{8}\frac{15x^3+25x-6}{\left(x^2+1\right)^2}+\frac{15}{8}\arctan{\left(x\right)}+C\\$

#128919 Całka 19

Napisane przez Mariusz M w 19.02.2017 - 23:34

$\sqrt{4-x^2}=\left(2+x\right)t\\\\\left(2+x\right)\left(2-x\right)=\left(2+x\right)^2t^2\\2-x=\left(2+x\right)t^2\\2-x=2t^2+xt^2\\x+xt^2=2-2t^2\\x\left(1+t^2\right)=2-2t^2\\x=\frac{2-2t^2}{1+t^2}=\frac{-2-2t^2+4}{1+t^2}=-2+\frac{4}{1+t^2}\\\\\left(2+x\right)t=\frac{4t}{1+t^2}\\\\\mbox{d}x=-4\left(1+t^2\right)^{-2}\cdot 2t\mbox{d}t\\\\\mbox{d}x=-\frac{8t}{\left(1+t^2\right)^{2}}\mbox{d}t\\\\\int{\frac{\left(1+t^2\right)^6}{64\left(1-t^2\right)^6}\cdot\frac{64t^3}{\left(1+t^2\right)^3}\cdot\left(-\frac{8t}{\left(1+t^2\right)^{2}}\right)\mbox{d}t}\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}\\\\\int{\left(-4t^3-4t^5\right)\frac{2t}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{4}{5}\int{\frac{3t^2+5t^4}{\left(1-t^2\right)^5}\mbox{d}t}\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{2}{5}\int{\left(3t+5t^3\right)\frac{2t}{\left(1-t^2\right)^5}\mbox{d}t}\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{1}{10}\frac{3t+5t^3}{\left(1-t^2\right)^4}-\frac{3}{10}\int{\frac{1+5t^2}{\left(1-t^2\right)^4}\mbox{d}t}\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{1}{10}\frac{3t+5t^3}{\left(1-t^2\right)^4}-\frac{3}{10}\int{\frac{1-t^2+6t^2}{\left(1-t^2\right)^4}\mbox{d}t}\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{1}{10}\frac{3t+5t^3}{\left(1-t^2\right)^4}-\frac{3}{10}\int{\frac{\mbox{d}t}{\left(1-t^2\right)^3}\mbox{d}t}-\frac{9}{10}\int{t\frac{2t}{\left(1-t^2\right)^4}\mbox{d}t}\\\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{1}{10}\frac{3t+5t^3}{\left(1-t^2\right)^4}-\frac{3}{10}\int{\frac{\mbox{d}t}{\left(1-t^2\right)^3}\mbox{d}t}-\frac{3}{10}\frac{t}{\left(1-t^2\right)^3}+\frac{3}{10}\int{\frac{\mbox{d}t}{\left(1-t^2\right)^3}}\\\\-8\int{\frac{t^4\left(1+t^2\right)}{\left(1-t^2\right)^6}\mbox{d}t}=-\frac{4}{5}\frac{t^3+t^5}{\left(1-t^2\right)^5}+\frac{1}{10}\frac{3t+5t^3}{\left(1-t^2\right)^4}-\frac{3}{10}\frac{t}{\left(1-t^2\right)^3}\\ \\$

#128916 Całka 37

Napisane przez Mariusz M w 19.02.2017 - 18:30

Jeśli chcesz sprowadzić do całki z funkcji wymiernej to

$\\-\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)\\\\\sqrt{-\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)}=\left(x+2+\sqrt{5}\right)t\\\\-\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=\left(x+2+\sqrt{5}\right)^2t^2\\\\-\left(x+2-\sqrt{5}\right)=\left(x+2+\sqrt{5}\right)t^2\\\\-x-2+\sqrt{5}=xt^2+\left(2+\sqrt{5}\right)t^2\\\\-x-xt^2=\left(2+\sqrt{5}\right)t^2+2-\sqrt{5}\\\\x\left(1+t^2\right)=-\left(2+\sqrt{5}\right)t^2-2+\sqrt{5}\\\\x=-\frac{\left(2+\sqrt{5}\right)t^2+2-\sqrt{5}}{1+t^2}=-\frac{\left(2+\sqrt{5}\right)t^2+2+\sqrt{5}-2\sqrt{5}}{1+t^2}=-2-\sqrt{5}+2\sqrt{5}\frac{1}{\left(1+t^2\right)}\\\\\left(x+2+\sqrt{5}\right)t=2\sqrt{5}\frac{t}{\left(1+t^2\right)}\\\\\mbox{d}x=-2\sqrt{5}\cdot\left(t^2+1\right)^{-2}\cdot 2t\mbox{d}t\\\\\mbox{d}x=-4\sqrt{5}\frac{t}{\left(t^2+1\right)^2}\mbox{d}t\\\\\int{\left(-\frac{\left(2+\sqrt{5}\right)t^2+2-\sqrt{5}}{1+t^2}-1\right)\cdot\frac{1+t^2}{2\sqrt{5}t}\cdot\left(-4\sqrt{5}\frac{t}{\left(t^2+1\right)^2}\right)\mbox{d}t}\\$

Jeśli nie chcesz sprowadzać całki do całki z funkcji wymiernej to

możesz zapisać trójmian kwadratowy pod pierwiastkiem w postaci kanonicznej

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