integral of (x+1)sec(x^2+2x)tan(x^2+2x)

Lösung

\int (x + 1) sec (x^{2} + 2 x) tan (x^{2} + 2 x) d x

\frac{1}{2} sec (x^{2} + 2 x) + C

Schritte zur Lösung

\int (x + 1) sec (x^{2} + 2 x) tan (x^{2} + 2 x) d x

Wende U-Substitution an

= \int u sec (u^{2} - 1) tan (u^{2} - 1) d u

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int sec (v) tan (v) d v

Wende U-Substitution an

= \frac{1}{2} \cdot \int 1 d w

Integral einer Konstanten :

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot w

Ersetze zurück

= \frac{1}{2} \cdot 1 \cdot sec ((x + 1)^{2} - 1)

Vereinfache

\frac{1}{2} \cdot 1 \cdot sec ((x + 1)^{2} - 1) : \frac{1}{2} sec (x^{2} + 2 x)

= \frac{1}{2} sec (x^{2} + 2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} sec (x^{2} + 2 x) + C

d er i v a t i v e sin^{3} (5 x)

y \to - 3 lim (4 y^{2} + 4 y + 1)

t an g e n t y = (2)^{x}, at x = 8

p a r t ia l f r a c t i o n \frac{x ^{2}}{x ^{2} + 1}

f^{^{'}} x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)