What is the integral of (z+1)e^{6z} ?

The integral of (z+1)e^{6z} is 1/6 e^{6z}z+5/36 e^{6z}+C

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integral of (z+1)e^{6z}

\int (z + 1) e^{6 z} d z

\frac{1}{6} e^{6 z} z + \frac{5}{36} e^{6 z} + C

Solution steps

\int (z + 1) e^{6 z} d z

Apply Integration By Parts

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

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

\int \frac{1}{6} e^{6 z} d z = \frac{1}{36} e^{6 z}

= \frac{1}{6} e^{6 z} (z + 1) - \frac{1}{36} e^{6 z}

Simplify

\frac{1}{6} e^{6 z} (z + 1) - \frac{1}{36} e^{6 z} : \frac{1}{6} e^{6 z} z + \frac{5}{36} e^{6 z}

= \frac{1}{6} e^{6 z} z + \frac{5}{36} e^{6 z}

Add a constant to the solution

= \frac{1}{6} e^{6 z} z + \frac{5}{36} e^{6 z} + C

\frac{d}{d x} (\frac{x ^{2} - 2 x + 2}{x - 1})

\int 1 + 4 y^{2} d y

\frac{ln ( x )}{x ^{3}}

y = \frac{5 x ^{2} + 3}{x ^{2} + 1}

\frac{\partial}{\partial x} (cos (\frac{x}{2}))

What is the integral of (z+1)e^{6z} ?
The integral of (z+1)e^{6z} is 1/6 e^{6z}z+5/36 e^{6z}+C