integral of sqrt(4x^2-5)

Lösung

\int 4 x^{2} - 5 d x

\frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

Schritte zur Lösung

\int 4 x^{2} - 5 d x

Trigonometrische Substitution anwenden

= \int \frac{5}{2} tan^{2} (u) sec (u) d u

Entferne die Konstante:

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

= \frac{5}{2} \cdot \int tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{5}{2} \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= \frac{5}{2} \cdot \int - sec (u) + sec^{3} (u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{5}{2} (- \int sec (u) d u + \int sec^{3} (u) d u)

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

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

= \frac{5}{2} (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arcsec (\frac{2}{5} x)

ein

= \frac{5}{2} (- ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x)) + \frac{1}{2} sec (arcsec (\frac{2}{5} x)) tan (arcsec (\frac{2}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x)))

Vereinfache

\frac{5}{2} (- ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x)) + \frac{1}{2} sec (arcsec (\frac{2}{5} x)) tan (arcsec (\frac{2}{5} x)) + \frac{1}{2} ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x))) : \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 x 4 x^{2} - 5 - 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

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