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I denne oppgaven regner vi ut det ubestemte integralet

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\\[ \\int (\\simplify{2{a}x+{b}})\\cdot e^{\\simplify{{a}x^2+{b}x+{c}}}dx\\]

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steg for steg ved bruk av substitusjon

", "advice": "

Først, bestemmer vi den kjernefunksjonen (indrefunksjonen) i integralet $y(x)=\\simplify{{a}x^2+{b}x+{c}}$

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Videre regner ut $y'(x)=\\dfrac{dy}{dx}=\\simplify{2{a}x+{b}}$

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Vi får en likning $\\dfrac{dy}{dx}=\\simplify{2{a}x+{b}}$ og fra denne likningen finner $dx=\\dfrac{dy}{\\simplify{2{a}x+{b}}}$

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Setter uttrykket for $y$ og $dx$ inn i integralet, forenkler og skriver integralet hvor integranden blir funksjon av $y$:

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\\[ \\int (\\simplify{2{a}x+{b}})\\cdot e^{\\simplify{{a}x^2+{b}x+{c}}}dx= \\int {(\\simplify{2{a}x+{b}})}\\cdot e^{y}\\dfrac{dy}{(\\simplify{2{a}x+{b}})}=\\int e^y dy\\]  

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Regner ut integralet

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\\[\\int e^y dy=e^y+C\\]  

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Og skriver inn tilbake uttrykk for $y(x)$:  

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\\[\\int e^y dy=e^y+C=e^{\\simplify{{a}x^2+{b}x+{c}}}+C\\]  

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Først, bestem kjernefunksjonen (indrefunksjonen) i integralet $y(x)=$ [[0]]

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Videre regn ut $y'(x)=\\dfrac{dy}{dx}=$ [[1]]. 

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Vi får en likning $\\dfrac{dy}{dx}=$ [[1]] og fra denne likningen finn $dx=$[[2]]

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Sett uttrykket for $y$ og $dx$ inn i integralet, forenkl og skriv integralet hvor integranden blir funksjon av $y$:

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$\\int$[[3]]$dy$    

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Regn ut integralet

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$\\int$[[3]]$dy$=[[4]]

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Og skriv inn tilbake uttrykk for $y(x)$:  [[5]]

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