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integral of sqrt(x^2+10x)

Lösung

\int x^{2} + 10 x d x

25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5})) + C

Schritte zur Lösung

\int x^{2} + 10 x d x

Vervollständige das Quadrat

x^{2} + 10 x : (x + 5)^{2} - 25

= \int (x + 5)^{2} - 25 d x

Wende U-Substitution an

= \int u^{2} - 25 d u

Trigonometrische Substitution anwenden

= \int 25 tan^{2} (v) sec (v) d v

Entferne die Konstante:

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

= 25 \cdot \int tan^{2} (v) sec (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 25 \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

Multipliziere aus

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

= 25 \cdot \int - sec (v) + sec^{3} (v) d v

Wende die Summenregel an:

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

= 25 (- \int sec (v) d v + \int sec^{3} (v) d v)

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

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

= 25 (- ln ∣ tan (v) + sec (v) ∣ + \frac{1}{2} sec (v) tan (v) + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= 25 (- ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} sec (arcsec (\frac{1}{5} (x + 5))) tan (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))))

Vereinfache

25 (- ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} sec (arcsec (\frac{1}{5} (x + 5))) tan (arcsec (\frac{1}{5} (x + 5))) + \frac{1}{2} ln tan (arcsec (\frac{1}{5} (x + 5))) + sec (arcsec (\frac{1}{5} (x + 5)))) : 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5}))

= 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5}))

Füge eine Konstante zur Lösung hinzu

= 25 (- ln (\frac{1}{5} x^{2} + 10 x + x + 5) + \frac{1}{50} (x + 5) x^{2} + 10 x + \frac{1}{2} ln (\frac{x ^{2} + 10 x + x + 5}{5})) + C

\int_{2}^{3} \frac{4}{3 - x} d x

\int \frac{2}{u ^{2}} d u

\frac{d}{d x} (arctan (sin (8 x)))

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

y^{^{'}} = e^{- y} (2 x - 4), y (0) = 1