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Example text

X - 3) 2 - 5 ' dx mit x - 3 = t, _ 1 '+ ar cosh ~) + C I) = I" + C1 Y t 2 - (\(5') 2' dt dx = dt. Mit der Substitution t '" dt = 1(5' cosh ='{5' sinh cp CD dCD ergibt sich das Integral fy 5 cosh 2 cp - 5' '{5'sinh =~ = fV 3x CD t x2 _ 8 ;2 3 Vx 2 , dx = 3 I dcp cosh cp - cp) + i V 2 - 5' - ~ In It x 3. (sinh CD + _ 6x + 4' - 2 ~ x2 - 8 f =5 c =~ Yt 2 'I sinh 2 cp dcp (_t_ '{5' J t: - 1 '_ ar cosh _t_ ) +C V5' - 5 + C1 ~ In Ix dx = 3 I - 3 + Yx 2 - 6x + 4 '1+ C1 d Vx 2 - 8 ' = 3 Yx 2 - 8 ' + C.

B stetige Funktionen erweitert werden, sofern f(x) dort keine reellen Nullstellen hat.

Lasung: Der Integrand ist unecht-gebrochen rational, muB also zunachst aufgespalten werden. Deshalb 1. Schritt: AusfUhrung der Division: 4 2 3 2) 5x 2 - x + 1 ( 2x - x - 5x + 1 ) : (x - x - 2x = 2x + 2 + -if--;,,--x 3 _ x 2 _ 2x 2. Schritt: Nullstellenbestimmung: x 3 - x 2 - 2x = 0 => xl = 0, x 2 = 2, x 3 _ x 2 _ 2x = x(x - 2)(x + 1). 2 Formale Integrationsmethoden 3. Schritt: Ansatz fUr Partialbruchzerlegung: A1 A2 A3 E-+--+-X x-2 x+1 und Koeffizientenbestimmung x = 0: x = 2: x = -1: 4. Schritt: Integration f 2X 4 - x 2 - 5x + 1 x 3 _x 2 _2x dx= f (2x+2)dx- = x 2 + 2x - 2.

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Calculus: all solutions by Stewart J.


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