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: 31 mar 2009, o 20:24
autor: Atraktor
\(\displaystyle{ \int \frac{5x^2 -12}{(x^2-6x+13)^2}}\)
ma ktoś jakiś pomysł?
obliczyć całkę
: 31 mar 2009, o 21:05
autor: jarzabek89
Rozkład na ułamki proste.
obliczyć całkę
: 31 mar 2009, o 21:19
autor: Atraktor
a jak mianownik rozłożyć? bo moim zdaniem nie da się
obliczyć całkę
: 31 mar 2009, o 21:22
autor: alia
Zobacz zasady rozkładu na ułamki proste
lub tutaj polecam twierdzenie 13.18.
obliczyć całkę
: 31 mar 2009, o 22:11
autor: Mariusz M
Atraktor pisze:a jak mianownik rozłożyć? bo moim zdaniem nie da się
Możesz użyć podstawienia
\(\displaystyle{ \frac{x-3}{2}=t}\)
a następnie skorzystać z
\(\displaystyle{ \int{ \frac{dx}{ \left( x^2+1\right)^n } dx}=}\)
\(\displaystyle{ \int{ \frac{dx}{\left(1+ x^2\right)^ {n-1} } }+ \int{ \frac{-x^2}{ \left(1+x^2 \right) ^n} }}\)
\(\displaystyle{ \int{ \frac{-x^2}{ \left(1+x^2 \right)^n } dx}= \frac{1}{2n-2} \int{ \frac{-\left( 2n-2\right)x*x }{ \left( 1+x^2\right)^n }dx}}\)
\(\displaystyle{ \frac{1}{2n-2} \frac{x}{ \left( 1+x^2\right) ^{n-1}}- \frac{1}{2n-2} \int{\frac{dx}{ \left( x^2+1\right)^{n-1} } dx}}\)
i ostatecznie
\(\displaystyle{ \int{ \frac{dx}{ \left( x^2+1\right)^n } dx}= \frac{1}{2n-2} \frac{x}{ \left( 1+x^2\right) ^{n-1}}- \frac{1}{2n-2} \int{\frac{dx}{ \left( x^2+1\right)^{n-1} } dx}+\int{ \frac{dx}{\left(1+ x^2\right)^ {n-1} } }}\)
\(\displaystyle{ \int{ \frac{dx}{ \left( 1+x^2\right)^n } }= \frac{1}{2n-2} \frac{x}{ \left( 1+x^2\right) ^{n-1}}+ \frac{2n-3}{2n-2} \int{\frac{dx}{ \left( 1+ x^2\right)^{n-1} } }}\)
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: 31 mar 2009, o 22:53
autor: Marcin_n
Rozłożenie na ułamki proste:
\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{Bx+C}{x ^{2}-6x+13}+ \frac{B _{2} x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)
\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{(Bx+C)(x ^{2}-6x+13)+B _{2}x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)
\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }=\frac{Bx ^{3}-6Bx ^{2}+13Bx+Cx ^{2}-6Cx+13C+B _{2}x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)
\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{Bx ^{3}+(C-6B)x ^{2}+(13B-6C+B _{2} )x +13C+C _{2} }{(x ^{2}-6x+13) ^{2}}}\)
\(\displaystyle{ (B=0 \wedge C-6B=5 \wedge i 13B-6C+B _{2}=0 \wedge 13C+C _{2}=-12) \Rightarrow B=0 \wedge C=5 \wedge B _{2}=30 \wedge C _{2} =-77 \Rightarrow}\)
\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{5}{x ^{2}-6x+13} + \frac{30x-77}{(x ^{2}-6x+13) ^{2}}}\)
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: 1 kwie 2009, o 01:36
autor: Mariusz M
Atraktor pisze:\(\displaystyle{ \int \frac{5x^2 -12}{(x^2-6x+13)^2}}\)
ma ktoś jakiś pomysł?
\(\displaystyle{ \int{ \frac{5x^2-12}{x^2-6x+13} dx}= \int{ \frac{5x^2-12}{ \left( \left( x-3\right)^2 +4\right)^2 } dx} =}\)
\(\displaystyle{ \int{ \frac{5x^2-12}{ 4^2\left( \left( \frac{x-3}{2} \right)^2 +1\right)^2 } dx} =}\)
\(\displaystyle{ \frac{1}{16} \int{ \frac{5x^2-12}{ \left( \left( \frac{x-3}{2} \right)^2 +1\right)^2 } dx} =}\)
\(\displaystyle{ t= \frac{x-3}{2}}\)
\(\displaystyle{ dt= \frac{1}{2}dx}\)
\(\displaystyle{ 2dt= dx}\)
\(\displaystyle{ x-3=2t}\)
\(\displaystyle{ x=2t+3}\)
\(\displaystyle{ \frac{2}{16} \int{ \frac{5 \left(2t+3 \right)^2-12 }{ \left(1+t^2 \right)^2 }dt }=}\)
\(\displaystyle{ \frac{1}{8} \int{ \frac{5 \left(4t^2+12t+9 \right)-12 }{ \left(1+t^2 \right)^2 }dt }=}\)
\(\displaystyle{ \frac{1}{8} \int{ \frac{\left(20t^2+60t+45 \right)-12 }{ \left(1+t^2 \right)^2 }dt }=}\)
\(\displaystyle{ \frac{1}{8} \int{ \frac{20t^2+60t+33}{ \left(1+t^2 \right)^2 }dt }=}\)
\(\displaystyle{ \frac{1}{8} \int{ \frac{20t^2+60t+33}{ \left(1+t^2 \right)^2 }dt }= \frac{1}{2}\left( \int{ \frac{20+20t^2}{ \left( 1+t^2\right) ^2} dt} -30 \int{ \frac{-2t}{\left(1+t^2\right)^2} +13 \int{ \frac{dt}{ \left( 1+t^2\right) ^2} } } \right) =}\)
\(\displaystyle{ \frac{1}{8} \left( 20\arctan{t}-30 \frac{1}{1+t^2}+13 \int{ \frac{dt}{ \left( 1+t^2\right)^2 } } \right)}\)
\(\displaystyle{ \int{ \frac{dt}{ \left( 1+t^2\right) ^2} }= \int{ \frac{dt}{1+t^2} }+ \int{ \frac{-t^2}{ \left( 1+t^2\right)^2 } }}\)
\(\displaystyle{ \int{ \frac{dt}{ \left( 1+t^2\right) ^2} }= \int{ \frac{dt}{1+t^2} }+ \frac{1}{2} \int{ \frac{-2t*tdt}{ \left( 1+t^2\right)^2 } }=}\)
\(\displaystyle{ \int{ \frac{dt}{ \left( 1+t^2\right) ^2} }= \int{ \frac{dt}{1+t^2} }+ \frac{1}{2} \left( \frac{t}{1+t^2} - \int{ \frac{dt}{1+t^2} } \right)}\)
\(\displaystyle{ \int{ \frac{dt}{ \left( 1+t^2\right) ^2} }= \frac{1}{2} \left( \frac{t}{1+t^2} + \int{ \frac{dt}{1+t^2} } \right)}\)
\(\displaystyle{ \int{ \frac{dt}{ \left( 1+t^2\right) ^2} }= \frac{1}{2} \left( \frac{t}{1+t^2} + \arctan{t} \right)}\)
\(\displaystyle{ = \frac{1}{8} \left( 20\arctan{t}-30 \frac{1}{1+t^2}+ \frac{13}{2} \left( \frac{t}{1+t^2} + \arctan{t} \right) \right)}\)
\(\displaystyle{ = \frac{1}{16} \left( 40\arctan{t}-60 \frac{1}{1+t^2}+ 13\left( \frac{t}{1+t^2} + \arctan{t} \right) \right)}\)
\(\displaystyle{ = \frac{1}{16} \left( 53\arctan{t}-60 \frac{1}{1+t^2}+ 13\frac{t}{1+t^2} \right)}\)
Powrót do zmiennej x
\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} - \frac{60}{1+ \left( \frac{x-3}{2} \right)^2 } } + \frac{ \frac{13}{2} \left( x-3\right) }{1+ \left( \frac{x-3}{4} \right)^2 } \right)}\)
\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} -60 \frac{1}{ \frac{4+x^2-6x+9}{4} } }+ \frac{ \frac{13}{2} \left( x-3\right) }{1+ \left( \frac{x^2-6x+9}{4}\right) } \right)}\)
\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2}} - \frac{240}{ x^2-6x+13 } + \frac{26 \left( x-3\right) }{x^2-6x+13 } \right)}\)
\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} - \frac{240}{ x^2-6x+13 } }+ \frac{26x-78}{x^2-6x+13 }\right)}\)
\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} + \frac{26x-318}{x^2-6x+13 }\right)+C}\)