Find the integral \( \int(z+1) e^{8 z} d z= \)

Calculate the integral \( \int(z+1)e^{8z}dz \). Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\int \left(z+1\right)e^{8z} dz\) - step1: Prepare for integration by parts: \(\begin{align}&u=z+1\\&dv=e^{8z}dz\end{align}\) - step2: Calculate the derivative: \(\begin{align}&du=dz\\&dv=e^{8z}dz\end{align}\) - step3: Evaluate the integral: \(\begin{align}&du=dz\\&v=\frac{e^{8z}}{8}\end{align}\) - step4: Substitute the values into formula: \(\left(z+1\right)\times \frac{e^{8z}}{8}-\int 1\times \frac{e^{8z}}{8} dz\) - step5: Calculate: \(\frac{\left(z+1\right)e^{8z}}{8}-\int \frac{e^{8z}}{8} dz\) - step6: Evaluate the integral: \(\frac{\left(z+1\right)e^{8z}}{8}-\frac{e^{8z}}{64}\) - step7: Simplify the expression: \(\frac{e^{8z}z}{8}+\frac{e^{8z}}{8}-\frac{e^{8z}}{64}\) - step8: Add the constant of integral C: \(\frac{e^{8z}z}{8}+\frac{e^{8z}}{8}-\frac{e^{8z}}{64} + C, C \in \mathbb{R}\) The integral of \( \int(z+1) e^{8z} dz \) is \( \frac{e^{8z}z}{8} + \frac{e^{8z}}{8} - \frac{e^{8z}}{64} + C \), where \( C \) is an arbitrary constant.