integral of sec(sqrt(x+1))tan(sqrt(x+1))

Lösung

\int sec (x + 1) tan (x + 1) d x

2 (x + 1 sec (x + 1) - ln tan (x + 1) + sec (x + 1)) + C

Schritte zur Lösung

\int sec (x + 1) tan (x + 1) d x

Wende U-Substitution an

= \int sec (u) tan (u) d u

Wende U-Substitution an

= \int sec (v) tan (v) \cdot 2 v d v

Entferne die Konstante:

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

= 2 \cdot \int sec (v) tan (v) v d v

Wende die partielle Integration an

= 2 (v sec (v) - \int sec (v) d v)

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

= 2 (v sec (v) - ln ∣ tan (v) + sec (v) ∣)

Ersetze zurück

= 2 (x + 1 sec (x + 1) - ln tan (x + 1) + sec (x + 1))

Füge eine Konstante zur Lösung hinzu

= 2 (x + 1 sec (x + 1) - ln tan (x + 1) + sec (x + 1)) + C

\int x^{2} (x^{3} + 1)^{\frac{1}{4}} d x

\frac{\partial}{\partial t} (2 cos (t))

f (x) = 2 arctan (x)

\frac{d}{d y} (\frac{1}{y})

\int - cos^{3} (x) d x