Hm, det var nog inte här jag parkera bilen

Conahd Saint:

Polynomdivision ger mig

x^3 - 13X^2 + 65X - 104 = (X^2 - 10X + 34) (X - 3) + X - 2

Och kvadratrotkomplettering i nämnaren ger mig

X^2 - 10X + 34 = (X + 5)^2 + 9

Men vart leder detta?

Gör ett försök:

Conahd Saint:

X^2 - 10X + 34 = (X + 5)^2 + 9

ska vara (x-5)^2 + 9

alltså

t^2(x-3)+x+7 / t^2+9

varför inte bara multiplicera allt med (t^2+9) och sedan substituera t tillbaka till (x-5). bör inte vara så svårt att göra en F(x) då.

Iavhé aka Mad World:

ska vara (x-5)^2 + 9

Sant.

Iavhé aka Mad World:

varför inte bara multiplicera allt med (t^2+9) och sedan substituera t tillbaka till (x-5). bör inte vara så svårt att göra en F(x) då.

Ska testa detta!

Conahd Saint:

Bestäm en primitiv funktion till

Conahd Saint:

f(x) = (x^3 - 13X^2 + 65X - 104)/(X^2 - 10X + 34)

Conahd Saint:

Fick tips om att använda polynomdivision, kvadratrotkomplettering i nämnaren och variabelbyte t = x - 5

Funktion:

f(x) = (x³-13x²+65x-104)/(x²-10x+34)

Söks:

∫(f(x)dx)

Förutsättningar:

D(f(x)+g(x)) = D(f(x))+D(g(x))
D(af(x)) = aD(f(x))
dz/dx = (dz/dy)dy/dx
D(x^a) = ax^(a-1)
D(ln(x)) = 1/x
D(ln(x²+1)) = 2x/(x²+1) = 2(x/(x²+1))
D(arctan(x)) = 1/(x²+1)
∫((x/(x²+1))dx) = (1/2)ln(x²+1)+A, A godtyckligt reellt tal
∫((1/(x²+1))dx) = arctan(x)+A, A godtyckligt reellt tal

Kvadratkomplettering:

x²-10x+34 =
x²+(-10x)+34 =
x²+(-10)x+34 =
x²+2(-5)x+34 =
x²+2x(-5)+34 =
x²+2x(-5)+(-5)²-(-5)²+34 =
(x+(-5))²-5²+34 =
(x-5)²-25+34 =
(x-5)²+9

Polynomdivision:

x-3
--------------------------------
x³-13x²+65x-104 | x²-10x+34
-(x³-10x²+34x)
--------------------------------
-3x²+31x-104
-(-3x²+30x-102)
--------------------------------
x-2

Kvot:

x-3

Rest:

x-2

Kontroll av polynomdivision:

(x²-10x+34)(x-3)+(x-2) =
x³-3x²-10x²+30x+34x-102+x-2 =
x³-13x²+65x-104

x³-13x²+65x-104 =
(x²-10x+34)(x-3)+(x-2) =
(x²-10x+34)(x-3)+x-2

f(x) =
(x³-13x²+65x-104)/(x²-10x+34) =
((x²-10x+34)(x-3)+(x-2))/(x²-10x+34) =
(x-3)+(x-2)/(x²-10x+34) =
x-3+(x-2)/(x²-10x+34) =
x-3+(x-2)/((x-5)²+9)

Substitution 1:

t = x-5
x = t+5

dt/dx = dx/dt = 1
dt = dx

∫(f(t)dt) = ∫(f(x-5)dx)
∫(f(x)dx) = ∫(f(t+5)dt)

f(x) =
f(t+5) =
(t+5)-3+((t+5)-2)/(t²+9) =
t+5-3+(t+5-2)/(t²+9) =
t+2+(t+3)/(t²+9) =
(t+2)+(t+3)/(t²+9)

Substitution 2:

3u = t
u = t/3 = (1/3)t
t = 3u

du/dt = 1/3
dt/du = 3
du = (1/3)dt
dt = 3du

∫(f(u)du) =
∫(f((1/3)t)((1/3)dt)) =
∫((1/3)(f((1/3)t)dt)) =
(1/3)∫(f((1/3)t)dt)
∫(f(t)dt) =
∫(f(3u)(3du)) =
∫(3(f(3u)du)) =
3∫(f(3u)du)

u = t/3 = (1/3)t = (x-5)/3 = (1/3)(x-5)
x = t+5 = 3u+5

du/dx = 1/3
dx/du = 3
du = (1/3)dx
dx = 3du

∫(f(u)du) =
∫(f((1/3)(x-5))((1/3)dx)) =
∫((1/3)(f((1/3)(x-5))dx)) =
(1/3)∫(f((1/3)(x-5))dx)
∫(f(x)dx) =
∫(f(3u+5)(3du)) =
∫(3(f(3u+5)du)) =
3∫(f(3u+5)du)

f(x) =
f(t+5) =
f(3u+5) =
3u+2+(3u+3)/((3u)²+9) =
3u+2+3(u+1)/((3u)²+9) =
3u+2+3(u+1)/(9u²+9) =
3u+2+3(u+1)/(9(u²+1)) =
3u+2+(3/9)(u+1)/(u²+1) =
3u+2+(1/3)(u+1)/(u²+1) =
3u+2+(1/3)u/(u²+1)+(1/3)/(u²+1) =
3u+2+(1/3)(u/(u²+1))+(1/3)(1/(u²+1))

Lösning:

∫(f(x)dx) =
∫(f(t+5)dt) =
3∫(f(3u+5)du) =
3∫((3u+2+(1/3)(u/(u²+1))+(1/3)(1/(u²+1)))du) =
3∫((3u)du)+3∫(2du)+3∫(((1/3)(u/(u²+1)))du)+3∫(((1/3)(1/(u²+1)))du) =
3∫(3((u)du))+3∫(2(du))+3∫((1/3)((u/(u²+1))du))+3∫((1/3)((1/(u²+1))du)) =
3*3∫((u)du)+3*2∫((1)du)+3(1/3)∫((u/(u²+1))du)+3(1/3)∫((1/(u²+1))du) =
9∫((u¹)du)+6∫((u^0)du)+(3*1/3)∫((u/(u²+1))du)+(3*1/3)∫((1/(u²+1))du) =
[ A godtyckligt reellt tal ]
9(1/(1+1))u^(1+1)+6(1/(0+1))u^(0+1)+(3/3)(1/2)ln(u²+1)+(3/3)arctan(u)+A =
9(1/2)u²+6(1/1)u¹+1(1/2)ln(u²+1)+1arctan(u)+A =
(9*1/2)u²+6*1u+(1*1/2)ln(u²+1)+arctan(u)+A =
(9/2)u²+6u+(1/2)ln(u²+1)+arctan(u)+A =
(9/2)((1/3)t)²+6(1/3)t+(1/2)ln(((1/3)t)²+1)+arctan((1/3)t)+A =
(9/2)(1/3)²t²+(6*1/3)t+(1/2)ln((1/3)²t²+1)+arctan((1/3)t)+A =
(9/2)(1²/3²)t²+(6/3)t+(1/2)ln((1²/3²)t²+1)+arctan((1/3)t)+A =
(9/2)(1/9)t²+2t+(1/2)ln((1/9)t²+1)+arctan((1/3)t)+A =
(9*1/(2*9))t²+2t+(1/2)ln((1/9)t²+9/9)+arctan((1/3)t)+A =
(9/18)t²+2t+(1/2)ln((1/9)t²+1*9/9)+arctan((1/3)t)+A =
(1/2)t²+2t+(1/2)ln((1/9)t²+(1/9)9)+arctan((1/3)t)+A =
(1/2)t²+2t+(1/2)ln((1/9)(t²+9))+arctan((1/3)t)+A =
(1/2)t²+2t+(1/2)(ln(1/9)+ln(t²+9))+arctan((1/3)t)+A =
(1/2)t²+2t+(1/2)ln(1/9)+(1/2)ln(t²+9)+arctan((1/3)t)+A =
(1/2)t²+2t+(1/2)ln(t²+9)+arctan((1/3)t)+A+(1/2)ln(1/9) =
(1/2)t²+2t+(1/2)ln(t²+9)+arctan((1/3)t)+(A+(1/2)ln(1/9)) =
[ B godtyckligt reellt tal
B = A+(1/2)ln(1/9) ]
(1/2)t²+2t+(1/2)ln(t²+9)+arctan((1/3)t)+B =
(1/2)(x-5)²+2(x-5)+(1/2)ln((x-5)²+9)+arctan((1/3)(x-5))+B =
(1/2)(x+(-5))²+2x-2*5+(1/2)ln((x+(-5))²+9)+arctan((1/3)(x-5))+B =
(1/2)(x²+2x(-5)+(-5)²)+2x-10+(1/2)ln(x²+2x(-5)+(-5)²+9)+arctan((1/3)(x-5))+B =
(1/2)(x²+2(-5)x+5²)+2x-10+(1/2)ln(x²+2(-5)x+5²+9)+arctan((1/3)(x-5))+B =
(1/2)(x²+(-10)x+25)+2x-10+(1/2)ln(x²+(-10)x+25+9)+arctan((1/3)(x-5))+B =
(1/2)(x²+(-10x)+25)+2x-10+(1/2)ln(x²+(-10x)+34)+arctan((1/3)(x-5))+B =
(1/2)(x²-10x+25)+2x-10+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-(1/2)10x+(1/2)25+2x-10+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-(1*10/2)x+1*25/2+2x-10+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-(10/2)x+25/2+2x-10+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-5x+2x+25*1/2-10*1+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+25(1/2)-10*2/2+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+25(1/2)-20/2+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+25(1/2)-20*1/2+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+25(1/2)-20(1/2)+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+(25-20)(1/2)+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+5(1/2)+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+5*1/2+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+5/2+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B =
(1/2)x²-3x+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+B+5/2 =
(1/2)x²-3x+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+(B+5/2) =
[ C godtyckligt reellt tal
C = B+5/2 ]
(1/2)x²-3x+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+C

Kontroll av resultat:

D(∫(f(x)dx)) =
D((1/2)x²-3x+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+C) =
D((1/2)x²+(-3x)+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+C) =
D((1/2)x²)+D(-3x)+D((1/2)ln(x²-10x+34))+D(arctan((1/3)(x-5)))+D(C) =
(1/2)D(x²)+D((-3)x)+(1/2)D(ln(x²-10x+34))+(1/(((1/3)(x-5))²+1))D((1/3)(x-5))+D(C1) =
(1/2)2x^(2-1)+(-3)D(x)+(1/2)(1/(x²-10x+34))D(x²-10x+34)+(1/3)(1/((1/3)²(x-5)²+1))D(x-5)+CD(1) =
(1*2/2)x¹+(-3)D(x¹)+(1/2)(1/(x²-10x+34))D(x²+(-10x)+34)+(1/3)(1/((1²/3²)(x+(-5))²+1))D(x+(-5))+CD(x^0) =
(2/2)x+(-3)1x^(1-1)+(1/2)(1/(x²-10x+34))(D(x²)+D(-10x)+D(34))+(1/3)(1/((1/9)(x²+2x(-5)+(-5)²)+1))(D(x)+D(-5))+C0x^(0-1) =
1x+(-3)x^0+(1/2)(1/(x²-10x+34))(2x^(2-1)+D((-10)x)+D(34*1))+(1/3)(1/((1/9)(x²+2(-5)x+5²)+9/9))(D(x¹)+D((-5)1))+0 =
x+(-3)1+(1/2)(1/(x²-10x+34))(2x¹+(-10)D(x)+34D(1))+(1/3)(1/((1/9)(x²+(-10)x+25)+1*9/9))(1x^(1-1)+(-5)D(1)) =
x+(-3)+(1/2)(1/(x²-10x+34))(2x+(-10)D(x¹)+34D(x^0))+(1/3)(1/((1/9)(x²+(-10x)+25)+(1/9)9))(x^0+(-5)D(x^0)) =
x-3+(1/2)(1/(x²-10x+34))(2x+(-10)1x^(1-1)+34*0x^(0-1))+(1/3)(1/((1/9)(x²-10x+25)+(1/9)9))(1+(-5)0x^(0-1)) =
x-3+(1/2)(1/(x²-10x+34))(2x+(-10)x^0+0)+(1/3)(1/((1/9)(x²-10x+25+9)))(1+0) =
x-3+(1/2)(1/(x²-10x+34))(2x+(-10)1)+(1/3)(1/(1/9))(1/(x²-10x+34))1 =
x-3+(1/2)(1/(x²-10x+34))(2x+(-10))+(1/3)9(1/(x²-10x+34)) =
x-3+(1/2)(1/(x²-10x+34))(2x-10)+(1*9/3)(1/(x²-10x+34)) =
x-3+(1/2)(1/(x²-10x+34))(2x-2*5)+(9/3)(1/(x²-10x+34)) =
x-3+(1/2)(1/(x²-10x+34))2(x-5)+3(1/(x²-10x+34)) =
x-3+(1*2/2)(x-5)(1/(x²-10x+34))+3(1/(x²-10x+34)) =
x-3+(2/2)(x-5)(1/(x²-10x+34))+3(1/(x²-10x+34)) =
x-3+1(x-5)(1/(x²-10x+34))+3(1/(x²-10x+34)) =
x-3+(x-5)(1/(x²-10x+34))+3(1/(x²-10x+34)) =
x-3+((x-5)+3)(1/(x²-10x+34)) =
x-3+(x-5+3)(1/(x²-10x+34)) =
x-3+(x-2)(1/(x²-10x+34)) =
x-3+(x-2)1/(x²-10x+34) =
x-3+(x-2)/(x²-10x+34) =
f(x)

Resultat:

∫(f(x)dx) = (1/2)x²-3x+(1/2)ln(x²-10x+34)+arctan((1/3)(x-5))+C
C godtyckligt reellt tal

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