What is the integral of (z+1)e^{9z} ?

The integral of (z+1)e^{9z} is 1/9 e^{9z}z+8/81 e^{9z}+C

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integral of (z+1)e^{9z}

\int (z + 1) e^{9 z} d z

\frac{1}{9} e^{9 z} z + \frac{8}{81} e^{9 z} + C

Solution steps

\int (z + 1) e^{9 z} d z

Apply Integration By Parts

: \frac{1}{9} e^{9 z} (z + 1) - \int \frac{1}{9} e^{9 z} d z

= \frac{1}{9} e^{9 z} (z + 1) - \int \frac{1}{9} e^{9 z} d z

\int \frac{1}{9} e^{9 z} d z = \frac{1}{81} e^{9 z}

= \frac{1}{9} e^{9 z} (z + 1) - \frac{1}{81} e^{9 z}

Simplify

\frac{1}{9} e^{9 z} (z + 1) - \frac{1}{81} e^{9 z} : \frac{1}{9} e^{9 z} z + \frac{8}{81} e^{9 z}

= \frac{1}{9} e^{9 z} z + \frac{8}{81} e^{9 z}

Add a constant to the solution

= \frac{1}{9} e^{9 z} z + \frac{8}{81} e^{9 z} + C

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What is the integral of (z+1)e^{9z} ?
The integral of (z+1)e^{9z} is 1/9 e^{9z}z+8/81 e^{9z}+C