integral of sqrt(x^2+4x+40)

Lösung

\int x^{2} + 4 x + 40 d x

36 (\frac{1}{72} (x + 2) x^{2} + 4 x + 40 + \frac{1}{2} ln (\frac{x + 2 + x ^{2} + 4 x + 40}{6})) + C

Schritte zur Lösung

\int x^{2} + 4 x + 40 d x

Vervollständige das Quadrat

x^{2} + 4 x + 40 : (x + 2)^{2} + 36

= \int (x + 2)^{2} + 36 d x

Wende U-Substitution an

= \int u^{2} + 36 d u

Trigonometrische Substitution anwenden

= \int 36 sec^{3} (v) d v

Entferne die Konstante:

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

= 36 \cdot \int sec^{3} (v) d v

Wende integrale Reduktion an

= 36 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

= 36 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= 36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x + 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x + 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2))))

Vereinfache

36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x + 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x + 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2)))) : 36 (\frac{1}{72} (x + 2) x^{2} + 4 x + 40 + \frac{1}{2} ln (\frac{x + 2 + x ^{2} + 4 x + 40}{6}))

= 36 (\frac{1}{72} (x + 2) x^{2} + 4 x + 40 + \frac{1}{2} ln (\frac{x + 2 + x ^{2} + 4 x + 40}{6}))

Füge eine Konstante zur Lösung hinzu

= 36 (\frac{1}{72} (x + 2) x^{2} + 4 x + 40 + \frac{1}{2} ln (\frac{x + 2 + x ^{2} + 4 x + 40}{6})) + C

d er i v a t i v e f (t) = 3 t sin (π t)

t an g e n t x^{3} + y^{3} = 7 x y, (2, - 4)

t an g e n t y = \frac{1}{x ^{2}}, (- 1, 1)

\frac{d}{d x} (\frac{ln ( 2 x )}{5 x})

\int cos^{2} (\frac{1}{2} x) d x