obliczyć całkę

[Atraktor]

\(\displaystyle{ \int \frac{5x^2 -12}{(x^2-6x+13)^2}}\)

ma ktoś jakiś pomysł?

[jarzabek89]

Rozkład na ułamki proste.

[Atraktor]

a jak mianownik rozłożyć? bo moim zdaniem nie da się

[alia]

Zobacz zasady rozkładu na ułamki proste
lub tutaj polecam twierdzenie 13.18.

[Mariusz M]

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

\(\displaystyle{ \int{ \frac{dx}{ \left( x^2+1\right)^n } dx}= \frac{1}{2n-2} \frac{x}{ \left( 1+x^2\right) ^{n-1}}- \frac{1}{2n-2} \int{\frac{dx}{ \left( x^2+1\right)^{n-1} } dx}+\int{ \frac{dx}{\left(1+ x^2\right)^ {n-1} } }}\)

\(\displaystyle{ \int{ \frac{dx}{ \left( 1+x^2\right)^n } }= \frac{1}{2n-2} \frac{x}{ \left( 1+x^2\right) ^{n-1}}+ \frac{2n-3}{2n-2} \int{\frac{dx}{ \left( 1+ x^2\right)^{n-1} } }}\)

[Marcin_n]

Rozłożenie na ułamki proste:

\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{Bx+C}{x ^{2}-6x+13}+ \frac{B _{2} x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)

\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{(Bx+C)(x ^{2}-6x+13)+B _{2}x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)

\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }=\frac{Bx ^{3}-6Bx ^{2}+13Bx+Cx ^{2}-6Cx+13C+B _{2}x+C _{2} }{(x ^{2}-6x+13) ^{2} }}\)

\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{Bx ^{3}+(C-6B)x ^{2}+(13B-6C+B _{2} )x +13C+C _{2} }{(x ^{2}-6x+13) ^{2}}}\)


\(\displaystyle{ (B=0 \wedge C-6B=5 \wedge i 13B-6C+B _{2}=0 \wedge 13C+C _{2}=-12) \Rightarrow B=0 \wedge C=5 \wedge B _{2}=30 \wedge C _{2} =-77 \Rightarrow}\)

\(\displaystyle{ \frac{5x ^{2}-12 }{(x ^{2}-6x+13) ^{2} }= \frac{5}{x ^{2}-6x+13} + \frac{30x-77}{(x ^{2}-6x+13) ^{2}}}\)

Dalej już łatwo obliczyć całkę

[Mariusz M]

\(\displaystyle{ \int \frac{5x^2 -12}{(x^2-6x+13)^2}}\)

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

\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} - \frac{60}{1+ \left( \frac{x-3}{2} \right)^2 } } + \frac{ \frac{13}{2} \left( x-3\right) }{1+ \left( \frac{x-3}{4} \right)^2 } \right)}\)

\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} -60 \frac{1}{ \frac{4+x^2-6x+9}{4} } }+ \frac{ \frac{13}{2} \left( x-3\right) }{1+ \left( \frac{x^2-6x+9}{4}\right) } \right)}\)

\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2}} - \frac{240}{ x^2-6x+13 } + \frac{26 \left( x-3\right) }{x^2-6x+13 } \right)}\)

\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} - \frac{240}{ x^2-6x+13 } }+ \frac{26x-78}{x^2-6x+13 }\right)}\)

\(\displaystyle{ \frac{1}{16} \left( 53\arctan{ \frac{x-3}{2} + \frac{26x-318}{x^2-6x+13 }\right)+C}\)