integral of (4e^z+15-1/(6z))

Lösung

\int (4 e^{z} + 15 - \frac{1}{6 z}) d z

4 e^{z} + 15 z - \frac{1}{6} ln ∣ z ∣ + C

Schritte zur Lösung

\int 4 e^{z} + 15 - \frac{1}{6 z} d z

Wende die Summenregel an:

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

= \int 4 e^{z} d z + \int 15 d z - \int \frac{1}{6 z} d z

\int 4 e^{z} d z = 4 e^{z}

\int 15 d z = 15 z

\int \frac{1}{6 z} d z = \frac{1}{6} ln ∣ z ∣

= 4 e^{z} + 15 z - \frac{1}{6} ln ∣ z ∣

Füge eine Konstante zur Lösung hinzu

= 4 e^{z} + 15 z - \frac{1}{6} ln ∣ z ∣ + C

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