What is the integral of (z+1)e^{8z} ?

The integral of (z+1)e^{8z} is 1/8 e^{8z}z+7/64 e^{8z}+C

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integral of (z+1)e^{8z}

\int (z + 1) e^{8 z} d z

\frac{1}{8} e^{8 z} z + \frac{7}{64} e^{8 z} + C

Solution steps

\int (z + 1) e^{8 z} d z

Apply Integration By Parts

= \frac{1}{8} e^{8 z} (z + 1) - \int \frac{1}{8} e^{8 z} d z

\int \frac{1}{8} e^{8 z} d z = \frac{1}{64} e^{8 z}

= \frac{1}{8} e^{8 z} (z + 1) - \frac{1}{64} e^{8 z}

Simplify

\frac{1}{8} e^{8 z} (z + 1) - \frac{1}{64} e^{8 z} : \frac{1}{8} e^{8 z} z + \frac{7}{64} e^{8 z}

= \frac{1}{8} e^{8 z} z + \frac{7}{64} e^{8 z}

Add a constant to the solution

= \frac{1}{8} e^{8 z} z + \frac{7}{64} e^{8 z} + C

What is the integral of (z+1)e^{8z} ?
The integral of (z+1)e^{8z} is 1/8 e^{8z}z+7/64 e^{8z}+C