integral of (e^{1-3x}+2cos(x/9))

Lösung

\int (e^{1 - 3 x} + 2 cos (\frac{x}{9})) d x

- \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9}) + C

Schritte zur Lösung

\int e^{1 - 3 x} + 2 cos (\frac{x}{9}) 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 e^{1 - 3 x} d x + \int 2 cos (\frac{x}{9}) d x

\int e^{1 - 3 x} d x = - \frac{1}{3} e^{1 - 3 x}

\int 2 cos (\frac{x}{9}) d x = 18 sin (\frac{x}{9})

= - \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9})

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{3} e^{1 - 3 x} + 18 sin (\frac{x}{9}) + C

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