What is the integral of (z+1)e^{3z} ?

The integral of (z+1)e^{3z} is 1/3 e^{3z}z+2/9 e^{3z}+C

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integral of (z+1)e^{3z}

\int (z + 1) e^{3 z} d z

\frac{1}{3} e^{3 z} z + \frac{2}{9} e^{3 z} + C

Solution steps

\int (z + 1) e^{3 z} d z

Apply Integration By Parts

: \frac{1}{3} e^{3 z} (z + 1) - \int \frac{1}{3} e^{3 z} d z

= \frac{1}{3} e^{3 z} (z + 1) - \int \frac{1}{3} e^{3 z} d z

\int \frac{1}{3} e^{3 z} d z = \frac{1}{9} e^{3 z}

= \frac{1}{3} e^{3 z} (z + 1) - \frac{1}{9} e^{3 z}

Simplify

\frac{1}{3} e^{3 z} (z + 1) - \frac{1}{9} e^{3 z} : \frac{1}{3} e^{3 z} z + \frac{2}{9} e^{3 z}

= \frac{1}{3} e^{3 z} z + \frac{2}{9} e^{3 z}

Add a constant to the solution

= \frac{1}{3} e^{3 z} z + \frac{2}{9} e^{3 z} + C

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\int 3 x (x - 2) d x

\int \frac{1}{x 2 - x ^{2}} d x

\int \frac{cos ( x )}{1 - cos ( x )} d x

2 x e^{2 x}

What is the integral of (z+1)e^{3z} ?
The integral of (z+1)e^{3z} is 1/3 e^{3z}z+2/9 e^{3z}+C