integral of (x^4)/(sqrt(x^2-9))

Lösung

\int \frac{x ^{4}}{x ^{2} - 9} d x

81 (\frac{1}{324} x^{3} x^{2} - 9 + \frac{1}{24} (x x^{2} - 9 + 9 ln (\frac{x + x ^{2} - 9}{3}))) + C

Schritte zur Lösung

\int \frac{x ^{4}}{x ^{2} - 9} d x

Vereinfache

x^{4} : x^{3} x

= \int \frac{x ^{3} x}{x ^{2} - 9} d x

Trigonometrische Substitution anwenden

= \int 81 sec^{5} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 81 \cdot \int sec^{5} (u) d u

Wende integrale Reduktion an

= 81 (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} \cdot \int sec^{3} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 81 (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Setze in

u = arcsec (\frac{1}{3} x)

ein

= 81 (\frac{sec ^{4} ( arcsec ( \frac{1}{3} x ) ) sin ( arcsec ( \frac{1}{3} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{3} x)) tan (arcsec (\frac{1}{3} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x))))

Vereinfache

81 (\frac{sec ^{4} ( arcsec ( \frac{1}{3} x ) ) sin ( arcsec ( \frac{1}{3} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arcsec (\frac{1}{3} x)) tan (arcsec (\frac{1}{3} x)) + \frac{1}{2} ln tan (arcsec (\frac{1}{3} x)) + sec (arcsec (\frac{1}{3} x)))) : 81 (\frac{1}{324} x^{3} x^{2} - 9 + \frac{1}{24} (x x^{2} - 9 + 9 ln (\frac{x + x ^{2} - 9}{3})))

= 81 (\frac{1}{324} x^{3} x^{2} - 9 + \frac{1}{24} (x x^{2} - 9 + 9 ln (\frac{x + x ^{2} - 9}{3})))

Füge eine Konstante zur Lösung hinzu

= 81 (\frac{1}{324} x^{3} x^{2} - 9 + \frac{1}{24} (x x^{2} - 9 + 9 ln (\frac{x + x ^{2} - 9}{3}))) + C

4 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\frac{d}{d x} (6 x e^{x})

t an g e n t f (x) = x^{3} - 3 x^{2} + 2 x - 2, at x = 0

\frac{\partial}{\partial y} (\frac{3 x - 4 y}{3 x + 4 y})

x \to 9 lim (cos (\frac{1}{x - 9}))