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Podstawienia Eulera


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#1 Mariusz M

Mariusz M

    Wielki Analityk

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Napisano 20.08.2015 - 08:37

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Najwyższa ocena

Dla tych którzy choć trochę znają rosyjski i chcą się dowiedzieć skąd się te podstawienia wzięły podaję odnośnik

 

http://mariuszm2011....l/Eulergeom.pdf
 

Dałem rosyjską wersję językową ponieważ językiem oryginału jest rosyjski

i jest ona wolna od ewentualnych błędów tłumaczy

Podstawienia Eulera sprowadzają całki postaci \int{R\left(x,\sqrt{ax^2+bx+c}\right)\mbox{d}x}

gdzie R\left(x,y\right) jest funkcją wymierną dwóch zmiennych do całek z funkcji wymiernej

 

Jeżeli a>0 to można zastosować pierwsze podstawienie Eulera

 

\int{R\left(x,\sqrt{ax^2+bx+c}\right)\mbox{d}x} \qquad a>0\\<br>\\\sqrt{ax^2+bx+c}=t-\sqrt{a}x\\<br>\\ax^2+bx+c=t^2-2\sqrt{a}tx+ax^2\\<br>\\bx+c=t^2-2\sqrt{a}tx\\<br>\\2\sqrt{a}tx+bx=t^2-c\\<br>\\x\left(2\sqrt{a}t+b\right)=t^2-c\\<br>\\x=\frac{t^2-c}{2\sqrt{a}t+b}\\<br>\\\sqrt{ax^2+bx+c}=t-\sqrt{a}x=\frac{2\sqrt{a}t^2+bt-\sqrt{a}t^2+\sqrt{a}c}{2\sqrt{a}t+b}\\<br>\\\sqrt{ax^2+bx+c}=\frac{\sqrt{a}t^2+bt+\sqrt{a}c}{2\sqrt{a}t+b}\\<br>\\\mbox{d}x=\frac{2t\left(2\sqrt{a}t+b\right)-2\sqrt{a}\left(t^2-c\right)}{\left(2\sqrt{a}t+b\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=2\frac{\sqrt{a}t^2+bt+\sqrt{a}c}{\left(2\sqrt{a}t+b\right)^2}\mbox{d}t\\<br>\\\int{R\left(\frac{t^2-c}{2\sqrt{a}t+b},\frac{\sqrt{a}t^2+bt+\sqrt{a}c}{2\sqrt{a}t+b}\right)\cdot 2\frac{\sqrt{a}t^2+bt+\sqrt{a}c}{\left(2\sqrt{a}t+b\right)^2}\mbox{d}t }\\<br>\\=\int{R_{1}\left(t\right)\mbox{d}t}\\<br>\\

 

Jeżeli c>0 to można zastosować drugie podstawienie Eulera

 

 

\int{R\left(x,\sqrt{ax^2+bx+c}\right)\mbox{d}x} \qquad c>0\\<br>\\\sqrt{ax^2+bx+c}=xt-\sqrt{c}\\<br>\\ax^2+bx+c=x^2t^2-2\sqrt{c}xt+c\\<br>\\ax^2+bx=x^2t^2-2\sqrt{c}xt\\<br>\\ax+b=xt^2-2\sqrt{c}t\\<br>\\2\sqrt{c}t+b=xt^2-ax\\<br>\\x\left(t^2-a\right)=2\sqrt{c}t+b\\<br>\\x=\frac{2\sqrt{c}t+b}{t^2-a}\\<br>\\\sqrt{ax^2+bx+c}=xt-\sqrt{c}=\frac{2\sqrt{c}t^2+bt-sqrt{c}t^2+a\sqrt{c}}{t^2-a}\\<br>\\\sqrt{ax^2+bx+c}=\frac{\sqrt{c}t^2+bt+a\sqrt{c}}{t^2-a}\\<br>\\\mbox{d}x=\frac{2\sqrt{c}\left(t^2-a\right)-2t\left(2\sqrt{c}t+b\right)}{\left(t^2-a\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=-2\frac{\sqrt{c}t^2+bt+a\sqrt{c}}{\left(t^2-a\right)^2}\mbox{d}t\\<br>\\\int{R\left(\frac{2\sqrt{c}t+b}{t^2-a},\frac{\sqrt{c}t^2+bt+a\sqrt{c}}{t^2-a}\right)\cdot \left(-2\right)\frac{\sqrt{c}t^2+bt+a\sqrt{c}}{\left(t^2-a\right)^2}\mbox{d}t}\\<br>\\=\int{R_{2}\left(t\right)\mbox{d}t}\\<br>\\

 

Jeżeli b^2-4ac>0 to można zastosować trzecie podstawienie Eulera

 

\int{R\left(x,\sqrt{ax^2+bx+c}\right)\mbox{d}x} \qquad b^2-4ac>0\\<br>\\\sqrt{a\left(x-\alpha\right)\left(x-\beta\right)}=\left(x-\alpha\right)t\\<br>\\a\left(x-\alpha\right)\left(x-\beta\right)=\left(x-\alpha\right)^2t^2\\<br>\\a\left(x-\beta\right)=\left(x-\alpha\right)t^2\\<br>\\ax-a\beta=xt^2-\alpha t^2\\<br>\\ax-xt^2=a\beta-\alpha t^2\\<br>\\x\left(a-t^2\right)=a\beta-\alpha t^2\\<br>\\x=\frac{a\beta-\alpha t^2}{a-t^2}=\alpha+a\frac{\beta-\alpha}{a-t^2}\\<br>\\\sqrt{a\left(x-\alpha\right)\left(x-\beta\right)}=\left(x-\alpha\right)t=a\frac{\left(\beta-\alpha\right)t}{a-t^2}\\<br>\\\mbox{d}x=-a\left(\beta-\alpha\right)\left(a-t^2\right)^{-2}\cdot \left(-2t\right)\mbox{d}t\\<br>\\\mbox{d}x=2a\left(\beta-\alpha\right)\frac{t}{\left(a-t^2\right)^{2}}\mbox{d}t\\<br>\\\int{R\left(\frac{a\beta-\alpha t^2}{a-t^2},a\frac{\left(\beta-\alpha\right)t}{a-t^2}\right)\cdot 2a\left(\beta-\alpha\right)\frac{t}{\left(a-t^2\right)^{2}}\mbox{d}t}\\<br>\\=\int{R_{3}\left(t\right)\mbox{d}t}\\<br>\\

 

 

Przykłady

 

</p>\\<p>\int{\frac{\mbox{d}x}{x\sqrt{x^2+4x-4}}}\\<br>\\\sqrt{x^2+4x-4}=t-x\\<br>\\x^2+4x-4=t^2-2tx+x^2\\<br>\\4x-4=t^2-2tx\\<br>\\2tx+4x=t^2+4\\<br>\\x\left(2t+4\right)=t^2+4\\<br>\\x=\frac{t^2+4}{2t+4}\\<br>\\t-x=\frac{2t^2+4t-t^2-4}{2t+4}=\frac{t^2+4t-4}{2t+4}\\<br>\\\mbox{d}x=\frac{2t\left(2t+4\right)-2\left(t^2+4\right)}{\left(2t+4\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+8t-8}{\left(2t+4\right)^2}\mbox{d}t\\<br>\\\int{\frac{2t+4}{t^2+4}\cdot\frac{2t+4}{t^2+4t-4}\cdot\frac{2\left(t^2+4t-4\right)}{\left(2t+4\right)^2}\mbox{d}t}\\<br>\\=2\int{\frac{\mbox{d}t}{t^2+4}}\\<br>\\=\frac{1}{2}\int{\frac{\mbox{d}t}{1+\left(\frac{t}{2}\right)^2}}\\<br>\\=\arctan{\left(\frac{x+\sqrt{x^2+4x-4}}{2}\right)}+C</p>\\<p>

 

</p>\\<p>\int{\frac{\mbox{d}x}{x\sqrt{2+x-x^2}}}=\sqrt{2}\int{\frac{\mbox{d}x}{x\sqrt{4+2x-2x^2}}}\\<br>\\\sqrt{4+2x-2x^2}=xt+2\\<br>\\4+2x-2x^2=x^2t^2+4xt+4\\<br>\\2-2x=xt^2+4t\\<br>\\xt^2+2x=2-4t\\<br>\\x\left(t^2+2\right)=2-4t\\<br>\\x=\frac{2-4t}{t^2+2}\\<br>\\xt+2=\frac{2t-4t^2+2t^2+4}{t^2+2}=\frac{-2\left(t^2-t-2\right)}{t^2+2}\\<br>\\\mbox{d}x=\frac{-4\left(t^2+2\right)-2t\left(2-4t\right)}{\left(t^2+2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{4t^2-4t-8}{\left(t^2+2\right)^2}\mbox{d}t\\<br>\\\sqrt{2}\int{\frac{t^2+2}{2-4t}\cdot\frac{t^2+2}{\left(-2\right)\left(t^2-t-2\right)}\cdot\frac{4\left(t^2-t-2\right)}{\left(t^2+2\right)^2}\mbox{d}t}\\<br>\\=\sqrt{2}\int{\frac{\mbox{d}t}{2t-1}}\\<br>\\=\frac{1}{\sqrt{2}}\int{\frac{\mbox{d}t}{2t-1}}\\<br>\\=\frac{1}{\sqrt{2}}\ln{\left|2t-1\right|}+C\\<br>\\=\frac{1}{\sqrt{2}}\ln{\left|\frac{2\sqrt{4+2x-2x^2}-x-4}{x}\right|}+C<br>\\

 

</p>\\<p>\int{\frac{\mbox{d}x}{\left(2x-3\right)\sqrt{4x-x^2}}}\\<br>\\\sqrt{4x-x^2}=xt\\<br>\\4x-x^2=x^2t^2\\<br>\\4-x=xt^2\\<br>\\4=xt^2+x\\<br>\\x\left(t^2+1\right)=4\\<br>\\x=\frac{4}{1+t^2}\\<br>\\\mbox{d}x=4\left(-1\right)\left(1+t^2\right)\cdot\left(2t\right)\mbox{d}t\\<br>\\\mbox{d}x=-\frac{8t}{\left(t^2+1\right)^2}\mbox{d}t\\<br>\\2x-3=\frac{8-3-3t^2}{1+t^2}=\frac{5-3t^2}{t^2+1}\\<br>\\\int{-\frac{t^2+1}{3t^2-5}\cdot\frac{t^2+1}{4t}\cdot\frac{\left(-8t\right)}{\left(t^2+1\right)^2}\mbox{d}t}\\<br>\\\int{\frac{2}{3t^2-5}\mbox{d}t}=\int{\frac{A}{\sqrt{3}t-\sqrt{5}}\mbox{d}t}+\int{\frac{B}{\sqrt{3}t+\sqrt{5}}\mbox{d}t}\\<br>\\A\left(\sqrt{3}t+\sqrt{5}\right)+B\left(\sqrt{3}t-\sqrt{5}\right)=2\\<br>\\\begin{cases}\sqrt{3}\left(A+B\right)=0\\\sqrt{5}\left(A-B\right)=2\end{cases}\\<br>\\\begin{cases}A+B=0\\A-B=\frac{2}{\sqrt{5}}\end{cases}\\<br>\\\begin{cases}B=-A\\2A=\frac{2}{\sqrt{5}}\end{cases}\\<br>\\\begin{cases}A=\frac{1}{\sqrt{5}}\\B=-\frac{1}{\sqrt{5}}\end{cases}\\<br>\\\int{\frac{2}{3t^2-5}\mbox{d}t}=\frac{1}{\sqrt{15}}\int{\frac{\sqrt{3}}{\sqrt{3}t-\sqrt{5}}\mbox{d}t}-\frac{1}{\sqrt{15}}\int{\frac{\sqrt{3}}{\sqrt{3}t+\sqrt{5}}\mbox{d}t}\\<br>\\\int{\frac{2}{3t^2-5}\mbox{d}t}=\frac{1}{\sqrt{15}}\ln{\left|\frac{\sqrt{3}t-\sqrt{5}}{\sqrt{3}t+\sqrt{5}}\right|}+C\\<br>\\\int{\frac{\mbox{d}x}{\left(2x-3\right)\sqrt{4x-x^2}}}=\frac{1}{\sqrt{15}}\ln{\left|\frac{\sqrt{3}\sqrt{4x-x^2}-sqrt{5}x}{\sqrt{3}\sqrt{4x-x^2}+\sqrt{5}x}\right|}+C<br>\\

 

<br>\\\int{\frac{\mbox{d}x}{x-\sqrt{x^2-x+1}}}\\<br>\\\sqrt{x^2-x+1}=t-x\\<br>\\x^2-x+1=t^2-2tx+x^2\\<br>\\-x+1=t^2-2tx\\<br>\\2tx-x=t^2-1\\<br>\\x\left(2t-1\right)=t^2-1\\<br>\\x=\frac{t^2-1}{2t-1}\\<br>\\x-\left(t-x\right)=2x-t\\<br>\\=\frac{2t^2-2-2t^2+t}{2t-1}=\frac{t-2}{2t-1}\\<br>\\\mbox{d}x=\frac{2t\left(2t-1\right)-2\left(t^2-1\right)}{\left(2t-1\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2-2t+2}{\left(2t-1\right)^2}\mbox{d}t\\<br>\\\int{\frac{2t-1}{t-2}\cdot\frac{2t^2-2t+2}{\left(2t-1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{2t^2-2t+2}{\left(t-2\right)\left(2t-1\right)}\mbox{d}t}\\<br>\\=\int{\frac{2t^2-2t+2}{2t^2-5t+2}\mbox{d}t}=\int{\frac{2t^2-5t+2+3t}{2t^2-5t+2}\mbox{d}t}\\<br>\\=\int{\mbox{d}t}+\int{\frac{3t}{\left(t-2\right)\left(2t-1\right)}\mbox{d}t}\\<br>\\=\int{\mbox{d}t}+\int{\frac{A}{t-2}\mbox{d}t}+\int{\frac{B}{2t-1}\mbox{d}t}\\<br>\\A\left(2t-1\right)+B\left(t-2\right)=3t\\<br>\\\begin{cases}2A+B=3\\-A-2B=0\end{cases}\\<br>\\\begin{cases}4A+2B=6\\-A-2B=0\end{cases}\\<br>\\\begin{cases}3A=6\\2B=-A\end{cases}\\<br>\\\begin{cases}A=2\\B=-1\end{cases}\\<br>\\\int{\frac{2t^2-2t+2}{\left(t-2\right)\left(2t-1\right)}\mbox{d}t}=\int{\mbox{d}t}+2\int{\frac{1}{t-2}\mbox{d}t}-\int{\frac{1}{2t-1}\mbox{d}t}\\<br>\\\int{\frac{2t^2-2t+2}{\left(t-2\right)\left(2t-1\right)}\mbox{d}t}=\int{\mbox{d}t}+2\int{\frac{1}{t-2}\mbox{d}t}-\frac{1}{2}\int{\frac{2}{2t-1}\mbox{d}t}\\<br>\\\int{\frac{2t^2-2t+2}{\left(t-2\right)\left(2t-1\right)}\mbox{d}t}=t+2\ln{\left|t-2\right|}-\frac{1}{2}\ln{\left|2t-1\right|}+C\\<br>\\\int{\frac{\mbox{d}x}{x-\sqrt{x^2-x+1}}}=x+\sqrt{x^2-x+1}+2\ln{\left|x-2+\sqrt{x^2-x+1}\right|}-\frac{1}{2}\ln{\left|2x-1+2\sqrt{x^2-x+1}\right|}+C\\<br>\\

 

<br>\\\int{\frac{\mbox{d}x}{x^2\left(x+\sqrt{1+x^2}\right)}}\\<br>\\\sqrt{1+x^2}=t-x\\<br>\\1+x^2=t^2-2tx+x^2\\<br>\\1=t^2-2tx\\<br>\\2tx=t^2-1\\<br>\\x=\frac{t^2-1}{2t}\\<br>\\\mbox{d}x=\frac{2t\cdot 2t-2\left(t^2-1\right)}{4t^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+2}{4t^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{t^2+1}{2t^2}\mbox{d}t\\<br>\\\int{\frac{4t^2}{\left(t^2-1\right)^2}\cdot\frac{1}{t}\cdot\frac{t^2+1}{2t^2}\mbox{d}t}\\<br>\\=\int{\frac{2t^2+2}{t\left(t^2-1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{2t}{\left(t^2-1\right)^2}\mbox{d}t}+\int{\frac{2}{t\left(t^2-1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{2t}{\left(t^2-1\right)^2}\mbox{d}t}+\int{\frac{2t^2-2\left(t^2-1\right)}{t\left(t^2-1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{4t}{\left(t^2-1\right)^2}\mbox{d}t}-\int{\frac{2}{t\left(t^2-1\right)}\mbox{d}t}\\<br>\\=\int{\frac{4t}{\left(t^2-1\right)^2}\mbox{d}t}+\int{\frac{2\left(t^2-1\right)-2t^2}{t\left(t^2-1\right)}\mbox{d}t}\\<br>\\=\int{\frac{4t}{\left(t^2-1\right)^2}\mbox{d}t}+2\int{\frac{\mbox{d}t}{t}}-\int{\frac{2t}{t^2-1}\mbox{d}t}\\<br>\\=-\frac{2}{t^2-1}+2\ln{\left|t\right|}-\ln{\left|t^2-1\right|}+C\\<br>\\=-\frac{2}{t^2-1}+\ln{\left|\frac{t^2}{t^2-1}\right|}+C\\<br>\\=-\frac{1}{x\left(x+\sqrt{1+x^2}\right)}+\ln{\left|\frac{x+\sqrt{1+x^2}}{x}\right|}+C<br>\\

 

<br>\\\int{\frac{x^2}{\sqrt{1-2x-x^2}}\mbox{d}x}\\<br>\\\sqrt{1-2x-x^2}=xt+1\\<br>\\1-2x-x^2=x^2t^2+2xt+1\\<br>\\-2-x=xt^2+2t\\<br>\\-2-2t=xt^2+x\\<br>\\x\left(1+t^2\right)=-2-2t^2\\<br>\\x=\frac{-2-2t}{1+t^2}\\<br>\\xt+1=\frac{-2t-2t^2+1+t^2}{1+t^2}=\frac{-t^2-2t+1}{1+t^2}\\<br>\\\mbox{d}x=\frac{-2\left(1+t^2\right)-2t\left(-2-2t\right)}{\left(1+t^2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+4t-2}{\left(1+t^2\right)^2}\mbox{d}t\\<br>\\-8\int{\frac{\left(t+1\right)^2}{\left(1+t^2\right)^2}\cdot\frac{1+t^2}{t^2+t-1}\cdot\frac{t^2+2t-1}{\left(1+t^2\right)^2}\mbox{d}t}\\<br>\\=-8\int{\frac{\left(1+t\right)^2}{\left(1+t^2\right)^3}\mbox{d}t}\\<br>\\=-8\left(\int{\frac{\mbox{d}t}{\left(1+t^2\right)^2}}+\int{\frac{2t}{\left(1+t^2\right)^3}\mbox{d}t}\right)\\<br>\\=-8\left(\int{\frac{\mbox{d}t}{\left(1+t^2\right)^2}}-\frac{1}{2}\int{\frac{\left(-4t\right)}{\left(1+t^2\right)^3}\mbox{d}t}\right)\\<br>\\=-8\left(\int{\frac{\mbox{d}t}{\left(1+t^2\right)}}+\int{\frac{t}{2}\cdot\frac{\left(-2t\right)}{\left(1+t^2\right)^2}}-\frac{1}{2}\int{\frac{\left(-4t\right)}{\left(1+t^2\right)^3}\mbox{d}t}\right)\\<br>\\=-8\left(\int{\frac{\mbox{d}t}{\left(1+t^2\right)}}+\frac{1}{2}\cdot\frac{t}{1+t^2}-\frac{1}{2}\int{\frac{\mbox{d}t}{1+t^2}}-\frac{1}{2}\int{\frac{\left(-4t\right)}{\left(1+t^2\right)^3}\mbox{d}t}\right)\\<br>\\=-8\left(\frac{1}{2}\cdot\frac{t}{1+t^2}+\frac{1}{2}\arctan{t}-\frac{1}{2}\cdot\frac{1}{\left(1+t^2\right)^2}\right)+C\\<br>\\=-4\left(\frac{t}{1+t^2}-\frac{1}{\left(1+t^2\right)^2}+\arctan{t}\right)+C\\<br>\\=4\left(\frac{1}{\left(1+t^2\right)^2}-\frac{t}{1+t^2}-\arctan{t}\right)+C\\<br>\\\int{\frac{x^2}{\sqrt{1-2x-x^2}}\mbox{d}x}=-\frac{1}{2}\left(x-3\right)\sqrt{1-2x-x^2}-4\arctan{\left(\frac{\sqrt{1-2x-x^2}-1}{x}\right)}+C<br>\\

 

<br>\\\int{\frac{3x^3-8x+5}{\sqrt{x^2-4x-7}}\mbox{d}x}\\<br>\\\sqrt{x^2-4x-7}=t-x\\<br>\\x^2-4x-7=t^2-2tx+x^2\\<br>\\-4x-7=t^2-2tx\\<br>\\2tx-4x=t^2+7\\<br>\\x\left(2t-4\right)=t^2+7\\<br>\\x=\frac{t^2+7}{2t-4}\\<br>\\t-x=\frac{2t^2-4t-t^2-7}{2t-4}\\<br>\\t-x=\frac{t^2-4t-7}{2t-4}\\<br>\\\mbox{d}x=\frac{2t\left(2t-4\right)-2\left(t^2+7\right)}{\left(2t-4\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2\left(t^2-4t-7\right)}{\left(2t-4\right)^2}\mbox{d}t\\<br>\\\int{\frac{3\left(t^2+7\right)^3-8\left(t^2+7\right)\left(2t-4\right)^2+5\left(2t-4\right)^3}{\left(2t-4\right)^3}\cdot\frac{2t-4}{t^2-4t-7}\cdot\frac{2\left(t^2-4t-7\right)}{\left(2t-4\right)^2}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{3\left(t^2+7\right)^3-8\left(t^2+7\right)\left(2t-4\right)^2+5\left(2t-4\right)^3}{\left(t-2\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{3t^6+31t^4+168t^3-151t^2+1376t-187}{\left(t-2\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{3\left(t-2\right)^6+36\left(t-2\right)^5+211\left(t-2\right)^4+896\left(t-2\right)^3+2321\left(t-2\right)^2+4356\left(t-2\right)+3993}{\left(t-2\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\left(3\int{\left(t-2\right)^2\mbox{d}t}+36\int{\left(t-2\right)\mbox{d}t}+211\int{\mbox{d}t}+896\int{\frac{\mbox{d}t}{t-2}}+2321\int{\frac{\mbox{d}t}{\left(t-2\right)^2}}+4356\int{\frac{\mbox{d}t}{\left(t-2\right)^3}}+3993\int{\frac{\mbox{d}t}{\left(t-2\right)^4}}\right)\\<br>\\=\frac{1}{8}\left(\left(t-2\right)^3+18\left(t-2\right)^2+211\left(t-2\right)-2321\cdot\frac{1}{t-2}-2178\cdot\frac{1}{\left(t-2\right)^2}-1331\cdot\frac{1}{\left(t-2\right)^3}+896\ln{\left|t-2\right|}\right)+C\\<br>\\\int{\frac{3x^3-8x+5}{\sqrt{x^2-4x-7}}\mbox{d}x}=\left(x^2+5x+36\right)\sqrt{x^2-4x-7}+112\cdot\ln{\left|x-2+\sqrt{x^2-4x-7}\right|}+C<br>\\

 

 

</p>\\<p>\int{\frac{x^4}{\sqrt{x^2+4x+5}}\mbox{d}x}\\<br>\\\sqrt{x^2+4x+5}=t-x\\<br>\\x^2+4x+5=t^2-2tx+x^2\\<br>\\4x+5=t^2-2tx\\<br>\\2tx+4x=t^2-5\\<br>\\x\left(2t+4\right)=t^2-5\\<br>\\x=\frac{t^2-5}{2t+4}\\<br>\\t-x=\frac{2t^2+4t-t^2+5}{2t+4}=\frac{t^2+4t+5}{2t+4}\\<br>\\\mbox{d}x=\frac{2t\left(2t+4\right)-2\left(t^2-5\right)}{\left(2t+4\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2\left(t^2+4t+5\right)}{\left(2t+4\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(t^2-5\right)^4}{\left(2t+4\right)^4}\cdot\frac{2t+4}{t^2+4t+5}\cdot\frac{2\left(t^2+4t+5\right)}{\left(2t+4\right)^2}\mbox{d}t}\\<br>\\=\frac{1}{16}\int{\frac{\left(t^2-5\right)^4}{\left(t+2\right)^{5}}\mbox{d}t}\\<br>\\=\frac{1}{16}\int{\frac{t^8-20t^6+150t^4-500t^2+625}{\left(t+2\right)^5}\mbox{d}t}\\<br>\\=\frac{1}{16}\int{\frac{\left(t+2\right)^8-16\left(t+2\right)^7+92\left(t+2\right)^6-208\left(t+2\right)^5+208\left(t+2\right)^3+92\left(t+2\right)^2+16\left(t+2\right)+1+70\left(t+2\right)^4}{\left(t+2\right)^5}\mbox{d}t}\\<br>\\=\frac{1}{16}\left(\int{\left(t+2\right)^3\mbox{d}t}-16\int{\left(t+2\right)^2\mbox{d}t}+92\int{\left(t+2\right)\mbox{d}t}-208\int{\mbox{d}t}+208\int{\frac{\mbox{d}t}{\left(t+2\right)^2}}+92\int{\frac{\mbox{d}t}{\left(t+2\right)^3}}+16\int{\frac{\mbox{d}t}{\left(t+2\right)^4}}+\int{\frac{\mbox{d}t}{\left(t+2\right)^5}}+70\int{\frac{\mbox{d}t}{t+2}}\right)\\<br>\\=\frac{1}{16}\left(\frac{1}{4}\left(t+2\right)^4-\frac{16}{3}\left(t+2\right)^3+46\left(t+2\right)^2-208\left(t+2\right)-208\cdot\frac{1}{t+2}-46\cdot\frac{1}{\left(t+2\right)^2}-\frac{16}{3}\cdot\frac{1}{\left(t+2\right)^3}-\frac{1}{4}\cdot\frac{1}{\left(t+2\right)^4}+70\ln{\left|t+2\right|}\right)+C\\<br>\\\int{\frac{x^4}{\sqrt{x^2+4x+5}}\mbox{d}x}=\frac{1}{24}\left(6x^3-28x^2+95x-290\right)\sqrt{x^2+4x+5}+\frac{35}{8}\cdot\ln{\left|x+2+sqrt{x^2+4x+5}\right|}+C<br>\\

 

</p>\\<p>\int{\frac{x^3-x+1}{\sqrt{x^2+2x+2}}\mbox{d}x}\\<br>\\\sqrt{x^2+2x+2}=t-x\\<br>\\x^2+2x+2=t^2-2tx+x^2\\<br>\\2x+2=t^2-2tx\\<br>\\2tx+2x=t^2-2\\<br>\\x\left(2t+2\right)=t^2-2\\<br>\\x=\frac{t^2-2}{2t+2}\\<br>\\t-x=\frac{2t^2+2t-t^2+2}{2t+2}=\frac{t^2+2t+2}{2t+2}\\<br>\\\mbox{d}x=\frac{2t\left(2t+2\right)-2\left(t^2-2\right)}{\left(2t+2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+4t+4}{\left(2t+4\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(t^2-2\right)^3-\left(t^2-2\right)\left(2t+2\right)^2+\left(2t+2\right)^3}{\left(2t+2\right)^3}\cdot\frac{2t+2}{t^2+2t+2}\cdot\frac{2t^2+4t+4}{\left(2t+4\right)^2}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{\left(t^2-2\right)^3-\left(t^2-2\right)\left(2t+2\right)^2+\left(2t+2\right)^3}{\left(t+1\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{t^6-10t^4+40t^2+40t+8}{\left(t+1\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\int{\frac{\left(t+1\right)^6-6\left(t+1\right)^5+5\left(t+1\right)^4+20\left(t+1\right)^3-5\left(t+1\right)^2-6\left(t+1\right)-1}{\left(t+1\right)^4}\mbox{d}t}\\<br>\\=\frac{1}{8}\left(\int{\left(t+1\right)^2\mbox{d}t}-6\int{\left(t+1\right)\mbox{d}t}+5\int{\mbox{d}t}-5\int{\frac{\mbox{d}t}{\left(t+1\right)^2}}-6\int{\frac{\mbox{d}t}{\left(t+1\right)^3}}-\int{\frac{\mbox{d}t}{\left(t+1\right)^4}}+20\int{\frac{\mbox{d}t}{t+1}}\right)\\<br>\\=\frac{1}{8}\left(\frac{1}{3}\left(t+1\right)^3-3\left(t+1\right)^2+5\left(t+1\right)+5\cdot\frac{1}{t+1}+3\cdot\frac{1}{\left(t+1\right)^2}+\frac{1}{3}\cdot\frac{1}{\left(t+1\right)^3}+20\ln{\left|t+1\right|}\right)+C\\<br>\\\int{\frac{x^3-x+1}{\sqrt{x^2+2x+2}}\mbox{d}x}=\frac{1}{6}\left(2x^2-5x+1\right)\sqrt{x^2+2x+2}+\frac{5}{2}\ln{\left|x+1+sqrt{x^2+2x+2}\right|}+C<br>\\

 

<br>\\\int{\frac{x^3}{\sqrt{1+2x-x^2}}\mbox{d}x}\\<br>\\\sqrt{1+2x-x^2}=xt+1\\<br>\\1+2x-x^2=x^2t^2+2tx+1\\<br>\\2-x=xt^2+2t\\<br>\\2-2t=xt^2+x\\<br>\\2-2t=x\left(t^2+1\right)\\<br>\\x=\frac{2-2t}{t^2+1}\\<br>\\xt+1=\frac{2t-2t^2+t^2+1}{t^2+1}\\<br>\\xt+1=-\frac{t^2-2t-1}{t^2+1}\\<br>\\\mbox{d}x=\frac{-2\left(t^2+1\right)-2t\left(2-2t\right)}{\left(t^2+1\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2-4t-2}{\left(t^2+1\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(2-2t\right)^3}{\left(1+t^2\right)^3}\cdot\frac{1+t^2}{-t^2+2t+1}\cdot\frac{2t^2-4t-2}{\left(t^2+1\right)^2}\mbox{d}t}\\<br>\\=16\int{\frac{\left(t-1\right)^3}{\left(1+t^2\right)^4}\mbox{d}t}\\<br>\\=16\int{\frac{\left(t-1\right)\left(t^2+1-2t\right)}{\left(1+t^2\right)^4}\mbox{d}t}\\<br>\\=16\left(\int{\frac{t-1}{\left(1+t^2\right)^3}\mbox{d}t}+\int{\frac{-2t^2-2+2t+2}{\left(1+t^2\right)^4}\mbox{d}t}\right)\\<br>\\=16\left(\int{\frac{t-3}{\left(1+t^2\right)^3}\mbox{d}t}+\int{\frac{2t+2}{\left(1+t^2\right)^4}\mbox{d}t}\right)\\<br>\\I_{n}=\int{\frac{\mbox{d}t}{\left(1+t^2\right)^n}}\\<br>\\I_{1}=\arctan{t}\\<br>\\I_{n}=\int{\frac{\mbox{d}t}{\left(1+t^2\right)^{n-1}}}-\int{\frac{t^2}{\left(1+t^2\right)^{n}}\mbox{d}t}\\<br>\\\begin{vmatrix}u=t&\mbox{d}u=\mbox{d}t\\\mbox{d}v=\frac{t}{\left(1+t^2\right)^n\mbox{d}t}&v=-\frac{1}{2n-2}\cdot\frac{1}{\left(1+t^2\right)^{n-1}}\end{vmatrix}\\<br>\\\int{\frac{t^2}{\left(1+t^2\right)^n}\mbox{d}t}=-\frac{1}{2n-1}\cdot\frac{x}{\left(x^2+1\right)^{n-1}}+\frac{1}{2n-2}\int{\frac{\mbox{d}t}{\left(1+t^2\right)^{n-1}}}\\<br>\\I_{n}=I_{n-1}+\frac{1}{2n-2}\cdot\frac{t}{\left(1+t^2\right)^{n-1}}-\frac{1}{2n-2}I_{n-1}\\<br>\\I_{n}=\frac{1}{2n-2}\cdot\frac{t}{\left(1+t^2\right)^{n-1}}+\frac{2n-3}{2n-2}I_{n-1}\\<br>\\\int{\frac{\mbox{d}t}{\left(1+t^2\right)^4}}=\frac{1}{6}\cdot\frac{t}{\left(1+t^2\right)^3}+\frac{5}{6}\int{\frac{\mbox{d}t}{\left(1+t^2\right)^3}}\\<br>\\\int{\frac{\mbox{d}t}{\left(1+t^2\right)^3}}=\frac{1}{4}\cdot\frac{t}{\left(1+t^2\right)^3}+\frac{3}{4}\int{\frac{\mbox{d}t}{\left(1+t^2\right)^2}}\\<br>\\\int{\frac{\mbox{d}t}{\left(1+t^2\right)^2}}=\frac{1}{2}\cdot\frac{t}{1+t^2}+\frac{1}{2}\arctan{t}\\<br>\\\int{\frac{\mbox{d}t}{\left(1+t^2\right)^3}}=\frac{1}{4}\cdot\frac{t}{\left(1+t^2\right)^2}+\frac{3}{8}\cdot\frac{t}{1+t^2}+\frac{3}{8}\arctan{t}\\<br>\\\int{\frac{\mbox{d}t}{\left(1+t^2\right)^4}}=\frac{1}{6}\cdot\frac{t}{\left(1+t^2\right)^3}+\frac{5}{24}\cdot\frac{t}{\left(1+t^2\right)^2}+\frac{5}{16}\cdot\frac{t}{1+t^2}+\frac{5}{16}\arctan{t}\\<br>\\16\left(\int{\frac{t-3}{\left(1+t^2\right)^3}\mbox{d}t}+\int{\frac{2t+2}{\left(1+t^2\right)^4}\mbox{d}t}\right)=16\left(-\frac{1}{4}\cdot\frac{1}{\left(1+t^2\right)^2}-3\left(\frac{1}{4}\cdot\frac{t}{\left(1+t^2\right)^2}+\frac{3}{8}\cdot\frac{t}{1+t^2}+\frac{3}{8}\arctan{t}\right)-\frac{1}{3}\cdot\frac{1}{\left(1+t^2\right)^3}+2\left(\frac{1}{6}\cdot\frac{t}{\left(1+t^2\right)^3}+\frac{5}{24}\cdot\frac{t}{\left(1+t^2\right)^2}+\frac{5}{16}\cdot\frac{t}{1+t^2}+\frac{5}{16}\arctan{t}\right)\right)+C\\<br>\\=16\left(\frac{1}{3}\cdot\frac{t-1}{\left(1+t^2\right)^3}-\frac{1}{12}\cdot\frac{4t+3}{\left(1+t^2\right)^2}-\frac{1}{2}\cdot\frac{t}{1+t^2}-\frac{1}{2}\arctan{t}\right)+C\\<br>\\\int{\frac{x^3}{\sqrt{1+2x-x^2}}\mbox{d}x}=-\frac{1}{6}\left(2x^2+5x+19\right)\sqrt{1+2x-x^2}-8\arctan{\left(\frac{\sqrt{1+2x-x^2}-1}{x}\right)}+C<br>\\

 

</p>\\<p>\int{\frac{\sqrt{1+x^2}}{2+x^2}\mbox{d}x}=\int{\frac{1+x^2}{\left(2+x^2\right)\sqrt{1+x^2}}\mbox{d}x}\\<br>\\=\int{\frac{\mbox{d}x}{\sqrt{1+x^2}}}-\int{\frac{\mbox{d}x}{\left(2+x^2\right)\sqrt{1+x^2}}}\\<br>\\\int{\frac{\mbox{d}x}{\sqrt{1+x^2}}}\\<br>\\\sqrt{1+x^2}=t-x\\<br>\\1+x^2=t^2-2tx+x^2\\<br>\\1=t^2-2tx\\<br>\\2tx=t^2-1\\<br>\\x=\frac{t^2-1}{2t}\\<br>\\\mbox{d}x=\frac{2t\cdot2t-2\left(t^2-1\right)}{4t^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{t^2+1}{2t^2}\mbox{d}t\\<br>\\\sqrt{1+x^2}=t-x=\frac{2t^2-\left(t^2-1\right)}{2t}\\<br>\\\sqrt{1+x^2}=\frac{t^2+1}{2t}\\<br>\\\int{\frac{2t}{t^2+1}\cdot\frac{t^2+1}{2t^2}\mbox{d}t}=\int{\frac{\mbox{d}t}{t}}\\<br>\\\int{\frac{\mbox{d}x}{\sqrt{1+x^2}}}=\ln{\left|x+sqrt{1+x^2}\right|}+C\\<br>\\\int{\frac{\mbox{d}x}{\left(2+x^2\right)\sqrt{1+x^2}}}\\<br>\\\sqrt{1+x^2}=xt+1\\<br>\\1+x^2=x^2t^2+2xt+1\\<br>\\x=xt^2+2t\\<br>\\x-xt^2=2t\\<br>\\x\left(1-t^2\right)=2t\\<br>\\x=\frac{2t}{1-t^2}\\<br>\\\sqrt{1+x^2}=xt+1=\frac{2t^2+1-t^2}{1-t^2}=\frac{1+t^2}{1-t^2}\\<br>\\2+x^2=2+\frac{4t^2}{\left(1-t^2\right)^2}=\frac{2-4t^2+2t^4+4t^2}{\left(1-t^2\right)^2}\\<br>\\2+x^2=\frac{2\left(1+t^4\right)}{\left(1-t^2\right)^2}\\<br>\\\mbox{d}x=\frac{2\left(1-t^2\right)+2t\cdot 2t}{\left(1-t^2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2+2t^2}{\left(1-t^2\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(1-t^2\right)^2}{2\left(1+t^4\right)}\cdot\frac{1-t^2}{1+t^2}\cdot\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{1-t^2}{1+t^4}\mbox{d}t}=\int{\frac{1-t^2}{\left(1-\sqrt{2}t+t^2\right)\left(1+\sqrt{2}t+t^2\right)}\mbox{d}t}\\<br>\\=\int{\frac{At+B}{1-\sqrt{2}t+t^2}\mbox{d}t}+\int{\frac{Ct+D}{1+\sqrt{2}t+t^2}\mbox{d}t}\\<br>\\\left(At+B\right)\left(1+\sqrt{2}t+t^2\right)+\left(Ct+D\right)\left(1-\sqrt{2}t+t^2\right)=1-t^2\\<br>\\\begin{cases}B+D=1\\A+\sqrt{2}B+C-\sqrt{2}D=0\\\sqrt{2}A+B-\sqrt{2}C+D=-1\\A+C=0\end{cases}\\<br>\\\begin{cases}A=-\frac{1}{\sqrt{2}}\\B=\frac{1}{2}\\C=\frac{1}{\sqrt{2}}\\D=\frac{1}{2}\end{cases}\\<br>\\-\frac{1}{2\sqrt{2}}\int{\frac{2t-\sqrt{2}}{1-\sqrt{2}t+t^2}\mbox{d}t}+\frac{1}{2\sqrt{2}}\int{\frac{2t+\sqrt{2}}{1+\sqrt{2}t+t^2}\mbox{d}t}\\<br>\\\int{\frac{1-t^2}{1+t^4}\mbox{d}t}=\frac{1}{2\sqrt{2}}\ln{\left|\frac{1+\sqrt{2}t+t^2}{1-\sqrt{2}t+t^2}\right|}+C\\<br>\\\int{\frac{\mbox{d}x}{\left(2+x^2\right)\sqrt{1+x^2}}}=\frac{1}{\sqrt{2}}\ln{\left|\frac{x+\sqrt{2+2x^2}}{\sqrt{2+x^2}}\right|}+C\\<br>\\\int{\frac{\sqrt{1+x^2}}{2+x^2}\mbox{d}x}=\ln{\left|x+sqrt{1+x^2}\right|}-\frac{1}{\sqrt{2}}\ln{\left|\frac{x+\sqrt{2+2x^2}}{\sqrt{2+x^2}}\right|}+C\\</p>\\<p>

 

</p>\\<p>\int{\frac{x-1}{x^2\sqrt{2x^2-2x+1}}\mbox{d}x}\\<br>\\\sqrt{2x^2-2x+1}=xt+1\\<br>\\2x^2-2x+1=x^2t^2+2xt+1\\<br>\\2x-2=xt^2+2t\\<br>\\2x-xt^2=2t+2\\<br>\\x\left(2-t^2\right)=2t+2\\<br>\\x=\frac{2t+2}{2-t^2}\\<br>\\x-1=\frac{2t+2-2+t^2}{2-t^2}=\frac{t^2+2t}{2-t^2}\\<br>\\xt+1=\frac{2t^2+2t+2-t^2}{2-t^2}=\frac{t^2+2t+2}{2-t^2}\\<br>\\\mbox{d}x=\frac{2\left(2-t^2\right)+2t\left(2t+2\right)}{\left(2-t^2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+4t+4}{\left(2-t^2\right)^2}\mbox{d}t\\<br>\\\int{\frac{t^2+2t}{2-t^2}\cdot\frac{\left(2-t^2\right)^2}{\left(2t+2\right)^2}\cdot\frac{2-t^2}{t^2+2t+2}\cdot\frac{2t^2+4t+4}{\left(2-t^2\right)^2}\mbox{d}t}\\<br>\\=\frac{1}{2}\int{\frac{t^2+2t}{\left(t+1\right)^2}\mbox{d}t}\\<br>\\=\frac{1}{2}\left(\int{\mbox{d}t}-\int{\frac{1}{\left(t+1\right)^2}\mbox{d}t}\right)\\<br>\\=\frac{1}{2}\left(\left(t+1\right)+\frac{1}{t+1}\right)+C\\<br>\\=\frac{t^2+2t+2}{2t+2}+C\\<br>\\=\frac{t^2+2t+2}{2-t^2}\cdot\frac{2-t^2}{2t+2}+C\\<br>\\\int{\frac{x-1}{x^2\sqrt{2x^2-2x+1}}\mbox{d}x}=\frac{\sqrt{2x^2-2x+1}}{x}+C</p>\\<p>

 

</p>\\<p>\int{\frac{\sqrt{x^2+2x+2}}{x^2}\mbox{d}x}\\<br>\\\sqrt{x^2+2x+2}=t-x\\<br>\\x^2+2x+2=t^2-2tx+x^2\\<br>\\2x+2=t^2-2tx\\<br>\\2tx+2x=t^2-2\\<br>\\x\left(2t+2\right)=t^2-2\\<br>\\x=\frac{t^2-2}{2t+2}\\<br>\\t-x=\frac{2t^2+2t-t^2+2}{2t+2}\\<br>\\\sqrt{x^2+2x+2}=\frac{t^2+2t+2}{2t+2}\\<br>\\\mbox{d}x=\frac{2t\left(2t+2\right)-2\left(t^2-2\right)}{\left(2t+2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+4t+4}{\left(2t+2\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(2t+2\right)^2}{\left(t^2-2\right)^2}\cdot \frac{t^2+2t+2}{2t+2}\cdot\frac{2t^2+4t+4}{\left(2t+2\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{\left(t^2+2t+2\right)^2}{\left(t+1\right)\left(t^2-2\right)^2}\mbox{d}t}\\<br>\\\int{\frac{\left(t^2+2t+2\right)^2}{\left(t+1\right)\left(t^2-2\right)^2}\mbox{d}t}=\int{\frac{A}{t+1}\mbox{d}t}+\int{\frac{B}{t-\sqrt{2}}\mbox{d}t}+\int{\frac{C}{\left(t-\sqrt{2}\right)^2}\mbox{d}t}+\int{\frac{D}{t+\sqrt{2}}\mbox{d}t}+\int{\frac{E}{\left(t+\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\A\left(t+\sqrt{2}\right)^2\left(t-\sqrt{2}\right)^2+B\left(t+1\right)\left(t-\sqrt{2}\right)\left(t+\sqrt{2}\right)^2+C\left(t+1\right)\left(t+\sqrt{2}\right)^2+D\left(t+1\right)\left(t+\sqrt{2}\right)\left(t-\sqrt{2}\right)^2+E\left(t+1\right)\left(t-\sqrt{2}\right)^2=\left(t^2+2t+2\right)^2\\<br>\\\begin{cases}A=1\\C=1+\sqrt{2}\\E=1-\sqrt{2}\\-B+D=-\sqrt{2}\\\left(1+sqrt{2}\right)B+\left(1-sqrt{2}\right)D=2\end{cases}\\<br>\\\begin{cases}A=1\\B=\frac{\sqrt{2}}{2}\\C=1+\sqrt{2}\\D=-\frac{\sqrt{2}}{2}\\E=1-\sqrt{2}\end{cases}\\<br>\\\int{\frac{1}{t+1}\mbox{d}t}+\frac{1}{\sqrt{2}}\int{\frac{1}{t-\sqrt{2}}\mbox{d}t}+\left(1+\sqrt{2}\right)\int{\frac{1}{\left(t-\sqrt{2}\right)^2}\mbox{d}t}-\frac{1}{\sqrt{2}}\int{\frac{1}{t+\sqrt{2}}\mbox{d}t}+\left(1-\sqrt{2}\right)\int{\frac{E}{\left(t+\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=\ln{\left|t+1\right|}+\frac{1}{\sqrt{2}}\ln{\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|}-\left(1+\sqrt{2}\right)\cdot\frac{1}{t-\sqrt{2}}-\left(1-\sqrt{2}\right)\cdot\frac{1}{t+\sqrt{2}}+C\\<br>\\=-\frac{2t+4}{t^2-2}+\ln{\left|t+1\right|}+\frac{1}{\sqrt{2}}\ln{\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|}+C\\<br>\\\int{\frac{\sqrt{x^2+2x+2}}{x^2}\mbox{d}x}=-\frac{\sqrt{x^2+2x+2}}{x}+\ln{\left|x+1+\sqrt{x^2+2x+2}\right|}+\frac{1}{\sqrt{2}}\ln{\left|\frac{x+2-\sqrt{2x^2+4x+4}}{x}\right|}+C\\</p>\\<p>

 

</p>\\<p>\int{\frac{x}{\left(x-1\right)^2\sqrt{1+2x-x^2}}\mbox{d}x}\\<br>\\\sqrt{1+2x-x^2}=xt+1\\<br>\\1+2x-x^2=x^2t^2+2xt+1\\<br>\\2-x=xt^2+2t\\<br>\\xt^2+x=2-2t\\<br>\\x\left(1+t^2\right)=2-2t\\<br>\\x=\frac{2-2t}{1+t^2}\\<br>\\x-1=\frac{-t^2-2t+1}{1+t^2}\\<br>\\xt+1=\frac{2t-2t^2+1+t^2}{1+t^2}\\<br>\\\sqrt{1+2x-x^2}=\frac{-t^2+2t+1}{1+t^2}\\<br>\\\mbox{d}x=\frac{-2\left(1+t^2\right)-2t\left(2-2t\right)}{\left(1+t^2\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2-4t-2}{\left(1+t^2\right)^2}\mbox{d}t\\<br>\\\int{\frac{2-2t}{1+t^2}\cdot\frac{\left(1+t^2\right)^2}{\left(t^2+2t-1\right)^2}\cdot\frac{1+t^2}{-t^2+2t+1}\cdot\frac{2t^2-4t-2}{\left(1+t^2\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{4t-4}{\left(t^2+2t-1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{4t+4}{\left(t^2+2t-1\right)^2}\mbox{d}t}-\int{\frac{8}{\left(\left(t+1\right)^2-2\right)^2}\mbox{d}t}\\<br>\\=-\frac{2}{t^2+2t-1}-\int{\frac{8}{\left(t+1-\sqrt{2}\right)^2\left(t+1+\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=-\frac{2}{t^2+2t-1}-\int{\frac{\left(t+1+\sqrt{2}\right)^2-2\left(t+1-\sqrt{2}\right)\left(t+1+\sqrt{2}\right)+\left(t+1-\sqrt{2}\right)^2}{\left(t+1-\sqrt{2}\right)^2\left(t+1+\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=-\frac{2}{t^2+2t-1}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}+2\int{\frac{1}{\left(t+1-\sqrt{2}\right)\left(t+1+\sqrt{2}\right)}\mbox{d}t}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=-\frac{2}{t^2+2t-1}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}+\frac{1}{\sqrt{2}}\int{\frac{\left(t+1+\sqrt{2}\right)-\left(t+1-\sqrt{2}\right)}{\left(t+1-\sqrt{2}\right)\left(t+1+\sqrt{2}\right)}\mbox{d}t}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=-\frac{2}{t^2+2t-1}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}+\frac{1}{\sqrt{2}}\int{\frac{1}{\left(t+1-\sqrt{2}\right)}\mbox{d}t}-\frac{1}{\sqrt{2}}\int{\frac{1}{\left(t+1+\sqrt{2}\right)}\mbox{d}t}-\int{\frac{1}{\left(t+1-\sqrt{2}\right)^2}\mbox{d}t}\\<br>\\=\frac{2t}{t^2+2t-1}+\frac{1}{\sqrt{2}}\ln{\left|\frac{t+1-\sqrt{2}}{t+1+\sqrt{2}}\right|}+C\\<br>\\\int{\frac{x}{\left(x-1\right)^2\sqrt{1+2x-x^2}}\mbox{d}x}=-\frac{1}{2}\cdot\frac{sqrt{1+2x-x^2}}{x-1}+\frac{1}{\sqrt{2}}\ln{\left|\frac{\sqrt{2}-\sqrt{1+2x-x^2}}{x-1}\right|}+C\\<br>\\

 

<br>\\\int{\frac{\mbox{d}x}{\left(x^2+x+1\right)\sqrt{x^2+x-1}}}\\<br>\\\sqrt{x^2+x-1}=t-x\\<br>\\x^2+x-1=t^2-2tx+x^2\\<br>\\x-1=t^2-2tx\\<br>\\2tx+x=t^2+1\\<br>\\x\left(2t+1\right)=t^2+1\\<br>\\x=\frac{t^2+1}{2t+1}\\<br>\\t-x=\frac{2t^2+t-t^2-1}{2t+1}\\<br>\\\sqrt{x^2+x-1}=\frac{t^2+t-1}{2t+1}\\<br>\\x^2+x+1=\frac{\left(t^2+t-1\right)^2+2\left(2t+1\right)^2}{\left(2t+1\right)^2}\\<br>\\x^2+x+1=\frac{t^4+2t^3+7t^2+6t+3}{\left(2t+1\right)^2}\\<br>\\\mbox{d}x=\frac{2t\left(2t+1\right)-2\left(t^2+1\right)}{\left(2t+1\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+2t-2}{\left(2t+1\right)^2}\mbox{d}t\\<br>\\\int{\frac{\left(2t+1\right)^2}{\left(t^4+2t^3+7t^2+6t+3\right)}\cdot\frac{2t+1}{t^2+t-1}\cdot\frac{2t^2+2t-2}{\left(2t+1\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{4t+2}{t^4+2t^3+7t^2+6t+3}\mbox{d}t}\\<br>\\t^4+2t^3+7t^2+6t+3\\<br>\\t^4+2t^3+t^2-\left(-6t^2-6t-3\right)\\<br>\\\left(t^2+t\right)^2-\left(-6t^2-6t-3\right)\\<br>\\\left(t^2+t+\frac{y}{2}\right)^2-\left(\left(y-6\right)t^2+\left(y-6\right)t+\frac{y^2}{4}-3\right)\\<br>\\y=6\\<br>\\\left(t^2+t+3\right)^2-6\\<br>\\\left(t^2+t+3-\sqrt{6}\right)\left(t^2+t+3+\sqrt{6}\right)\\<br>\\\int{\frac{4t+2}{\left(t^2+t+3-\sqrt{6}\right)\left(t^2+t+3+\sqrt{6}\right)}\mbox{d}t}\\<br>\\\int{\frac{4t+2}{t^4+2t^3+7t^2+6t+3}\mbox{d}t}=\int{\frac{At+B}{\left(t^2+t+3-\sqrt{6}\right)}\mbox{d}t}+\int{\frac{Ct+D}{\left(t^2+t+3+\sqrt{6}\right)}\mbox{d}t}\\<br>\\\left(At+B\right)\left(t^2+t+3+\sqrt{6}\right)+\left(Ct+D\right)\left(t^2+t+3-\sqrt{6}\right)=4t+2\\<br>\\\begin{cases}A+C=0\\A+B+C+D=0\\\left(3+\sqrt{6}\right)A+B+\left(3-\sqrt{6}\right)C+D=4\\\left(3+\sqrt{6}\right)B+\left(3-\sqrt{6}\right)D=2\end{cases}\\<br>\\\begin{cases}A=\frac{2}{\sqrt{6}}\\B=\frac{1}{\sqrt{6}}\\C=-\frac{2}{\sqrt{6}}\\D=-\frac{1}{\sqrt{6}}\end{cases}\\<br>\\\int{\frac{4t+2}{t^4+2t^3+7t^2+6t+3}\mbox{d}t}=\frac{1}{\sqrt{6}}\int{\frac{2t+1}{\left(t^2+t+3-\sqrt{6}\right)}\mbox{d}t}-\frac{1}{\sqrt{6}}\int{\frac{2t+1}{\left(t^2+t+3+\sqrt{6}\right)}\mbox{d}t}\\<br>\\=\frac{1}{\sqrt{6}}\ln{\left|\frac{t^2+t+3-\sqrt{6}}{t^2+t+3+\sqrt{6}}\right|}+C\\<br>\\\int{\frac{\mbox{d}x}{\left(x^2+x+1\right)\sqrt{x^2+x-1}}}=\frac{2}{\sqrt{6}}\ln{\left|\frac{\sqrt{2}\left(2x+1\right)+\sqrt{3}\cdot\sqrt{x^2+x-1}}{\sqrt{x^2+x+1}}\right|}+C<br>\\

 

</p>\\<p>\int{\frac{x-\sqrt{x^2+3x+2}}{x+\sqrt{x^2+3x+2}}\mbox{d}x}\\<br>\\t=x+\sqrt{x^2+3x+2}\\<br>\\x-\sqrt{x^2+3x+2}=x-\left(t-x\right)=2x-t\\<br>\\t=x+\sqrt{x^2+3x+2}\\<br>\\\sqrt{x^2+3x+2}=t-x\\<br>\\x^2+3x+2=t^2-2tx+x^2\\<br>\\3x+2=t^2-2tx\\<br>\\2tx+3x=t^2-2\\<br>\\x\left(2t+3\right)=t^2-2\\<br>\\x=\frac{t^2-2}{2t+3}\\<br>\\2x-t=\frac{2t^2-4-2t^2-3t}{2t+3}\\<br>\\x-\sqrt{x^2+3x+2}=-\frac{3t+4}{2t+3}\\<br>\\\mbox{d}x=\frac{2t\left(2t+3\right)-2\left(t^2-2\right)}{\left(2t+3\right)^2}\mbox{d}t\\<br>\\\mbox{d}x=\frac{2t^2+6t+4}{\left(2t+3\right)^2}\mbox{d}t\\<br>\\-\int{\frac{3t+4}{2t+3}\cdot\frac{1}{t}\cdot\frac{2t^2+6t+4}{\left(2t+3\right)^2}\mbox{d}t}\\<br>\\=\int{\frac{-6t^3-26t^2-36t-16}{t\left(2t+3\right)^3}\mbox{d}t}\\<br>\\=\int{\frac{A}{t}\mbox{d}t}+\int{\frac{B}{2t+3}\mbox{d}t}+\int{\frac{C}{\left(2t+3\right)^2}\mbox{d}t}+\int{\frac{D}{\left(2t+3\right)^3}\mbox{d}t}\\<br>\\A\left(2t+3\right)^3+Bt\left(2t+3\right)^2+Ct\left(2t+3\right)+Dt=-6t^3-26t^2-36t-16\\<br>\\\begin{cases}8A+4B=-6\\36A+12B+2C=-26\\54A+9B+3C+D=-36\\27A=-16\end{cases}\\<br>\\\begin{cases}A=-\frac{16}{27}\\B=-\frac{17}{54}\\C=-\frac{4}{9}\\D=\frac{1}{6}\end{cases}\\<br>\\\int{\frac{-6t^3-26t^2-36t-16}{t\left(2t+3\right)^3}\mbox{d}t}=-\frac{16}{27}\int{\frac{\mbox{d}t}{t}}-\frac{17}{108}\int{\frac{2\mbox{d}t}{2t+3}}+\frac{2}{9}\int{\frac{\left(-2\right)\mbox{d}t}{\left(2t+3\right)^2}}-\frac{1}{24}\int{\frac{\left(-4\right)\mbox{d}t}{\left(2t+3\right)^3}}\\<br>\\=\frac{2}{9}\cdot\frac{1}{2t+3}-\frac{1}{24}\cdot\frac{1}{\left(2t+3\right)^2}-\frac{16}{27}\ln{\left|t\right|}-\frac{17}{108}\ln{\left|2t+3\right|}+C\\<br>\\\int{\frac{x-\sqrt{x^2+3x+2}}{x+\sqrt{x^2+3x+2}}\mbox{d}x}=-\frac{1}{3}x^2-\frac{5}{9}x+\frac{1}{18}\left(6x+1\right)\sqrt{x^2+3x+2}-\frac{16}{27}\ln{\left|x+sqrt{x^2+3x+2}\right|}-\frac{17}{108}\ln{\left|2x+3+2\sqrt{x^2+3x+2}\right|}+C</p>\\<p>

 

\int{\frac{\mbox{d}x}{x^2\left(4x^2-3\right)^2\sqrt{x^2-1}}}\\</p>\\<p>\sqrt{x^2-1}=t-x\\</p>\\<p>x^2-1=t^2-2tx+x^2\\</p>\\<p>-1=t^2-2tx\\</p>\\<p>2tx=t^2+1\\</p>\\<p>x=\frac{t^2+1}{2t}\\</p>\\<p>\mbox{d}x=\frac{2t\cdot2t-2\left(t^2+1\right)}{4t^2}\mbox{d}t\\</p>\\<p>\mbox{d}x=\frac{t^2-1}{2t^2}\mbox{d}t\\</p>\\<p>\sqrt{x^2-1}=t-x=\frac{2t^2-t^2-1}{2t}=\frac{t^2-1}{2t}\\</p>\\<p>4x^2-3=\frac{\left(t^2+1\right)^2-3t^2}{t^2}=\frac{t^4-t^2+1}{t^2}\\</p>\\<p>\int{\frac{4t^2}{\left(t^2+1\right)^2}\cdot\frac{t^4}{\left(t^4-t^2+1\right)^2}\cdot\frac{2t}{t^2-1}\cdot\frac{t^2-1}{2t^2}\mbox{d}t}\\</p>\\<p>\int{\frac{4t^5}{\left(t^2+1\right)^2\left(t^4-t^2+1\right)^2}\mbox{d}t}\\</p>\\<p>=\frac{2}{3}\int{\frac{6t^5}{\left(t^6+1\right)^2}\mbox{d}t}\\</p>\\<p>=-\frac{2}{3}\cdot\frac{1}{t^6+1}+C\\</p>\\<p>=-\frac{2}{3}\cdot\frac{1}{\left(x+\sqrt{x^2-1}\right)^6+1}+C</p>\\<p>


Użytkownik Mariusz M edytował ten post 12.09.2015 - 01:58

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