integral of (sin(4x)+cosh(3x))

Lösung

\int (sin (4 x) + cosh (3 x)) d x

- \frac{1}{4} cos (4 x) + \frac{1}{3} sinh (3 x) + C

Schritte zur Lösung

\int sin (4 x) + cosh (3 x) d x

Wende die Summenregel an:

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

= \int sin (4 x) d x + \int cosh (3 x) d x

\int sin (4 x) d x = - \frac{1}{4} cos (4 x)

\int cosh (3 x) d x = \frac{1}{3} sinh (3 x)

= - \frac{1}{4} cos (4 x) + \frac{1}{3} sinh (3 x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} cos (4 x) + \frac{1}{3} sinh (3 x) + C

\int_{- 1}^{3} (2 x^{2} - x + 5) d x

t an g e n t y = 3 x + 1, (8, 5)

\int 2 x^{2} - 6 x d x

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

d er i v a t i v e y = - 2 e^{5 x^{2}}