Answer
$= -\frac{z}{\ln 10\times10^{z}} - \frac{1}{(\ln 10)^{2}\times 10^{z}} + C$
Work Step by Step
$\int \frac{z}{10^{z}}dz$
$\int z(10)^{-z}dz$
$u = z$
$u' = 1$
$\frac{du}{dz} = 1$
$du = 1 dz$
$dv = 10^{-z}dz$
$dv = e^{\ln (10^{-z})}$
$dv = e^{-z \ln 10}$
$v = \frac{1}{-\ln 10}e^{-z \ln 10}$
$v = \frac{1}{-\ln 10}10^{-z}$
$v = -\frac{1}{\ln 10\times10^{z}}$
$uv - \int vdu$
$= (z)(-\frac{1}{\ln 10\times10^{z}}) - \int (-\frac{1}{\ln 10\times10^{z}}) dz$
$= (z)(-\frac{1}{\ln 10\times10^{z}}) - \int \frac{1}{-\ln 10}10^{-z} dz$
$= (z)(-\frac{1}{\ln 10\times10^{z}}) + \frac{1}{\ln 10}\int 10^{-z} dz$
$= (z)(-\frac{1}{\ln 10\times10^{z}}) - \frac{1}{\ln 10}(\frac{1}{\ln 10})10^{-z} + C$
$= (z)(-\frac{1}{\ln 10\times10^{z}}) - \frac{1}{(\ln 10)^{2}}10^{-z} + C$
$= -\frac{z}{\ln 10\times10^{z}} - \frac{1}{(\ln 10)^{2}\times 10^{z}} + C$
