What is the integral of (4e^z+15-1/(6z)) ?

The integral of (4e^z+15-1/(6z)) is 4e^z+15z-1/6 ln|z|+C

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

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

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

Solution steps

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

Apply the Sum Rule:

\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 ∣

Add a constant to the solution

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

What is the integral of (4e^z+15-1/(6z)) ?
The integral of (4e^z+15-1/(6z)) is 4e^z+15z-1/6 ln|z|+C