integral of (x^2)/(sqrt(x^2-16))

Lösung

\int \frac{x ^{2}}{x ^{2} - 16} d x

16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4})) + C

Schritte zur Lösung

\int \frac{x ^{2}}{x ^{2} - 16} d x

Trigonometrische Substitution anwenden

= \int 16 sec^{3} (u) d u

Entferne die Konstante:

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

= 16 \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 16 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 16 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arcsec (\frac{1}{4} x)

ein

= 16 (\frac{sec ^{2} ( arcsec ( \frac{1}{4} x ) ) sin ( arcsec ( \frac{1}{4} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x)))

Vereinfache

16 (\frac{sec ^{2} ( arcsec ( \frac{1}{4} x ) ) sin ( arcsec ( \frac{1}{4} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{1}{4} x)) + sec (arcsec (\frac{1}{4} x))) : 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4}))

= 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4}))

Füge eine Konstante zur Lösung hinzu

= 16 (\frac{1}{32} x x^{2} - 16 + \frac{1}{2} ln (\frac{x ^{2} - 16 + x}{4})) + C

\int_{0}^{0.5} (300 x) d x

y^{^{''}} = 4 t^{3} + t

a re a y = (x - 5) x (x + 5), y = 5 - x

in v erse l a pl a ce \frac{( 1 - e ^{- s} )}{s}

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