What is the integral of (z+6)e^{4z} ?

The integral of (z+6)e^{4z} is 1/4 e^{4z}z+23/16 e^{4z}+C

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integral of (z+6)e^{4z}

\int (z + 6) e^{4 z} d z

\frac{1}{4} e^{4 z} z + \frac{23}{16} e^{4 z} + C

Solution steps

\int (z + 6) e^{4 z} d z

Apply Integration By Parts

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

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

\int \frac{1}{4} e^{4 z} d z = \frac{1}{16} e^{4 z}

= \frac{1}{4} e^{4 z} (z + 6) - \frac{1}{16} e^{4 z}

Simplify

\frac{1}{4} e^{4 z} (z + 6) - \frac{1}{16} e^{4 z} : \frac{1}{4} e^{4 z} z + \frac{23}{16} e^{4 z}

= \frac{1}{4} e^{4 z} z + \frac{23}{16} e^{4 z}

Add a constant to the solution

= \frac{1}{4} e^{4 z} z + \frac{23}{16} e^{4 z} + C

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What is the integral of (z+6)e^{4z} ?
The integral of (z+6)e^{4z} is 1/4 e^{4z}z+23/16 e^{4z}+C