What is the integral of z/(10^z) ?

The integral of z/(10^z) is -(10^{-z}z)/(ln(10))-1/(ln^2(10))*10^{-z}+C

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integral of z/(10^z)

\int \frac{z}{1 0 ^{z}} d z

- \frac{1 0 ^{- z} z}{ln ( 10 )} - \frac{1}{ln ^{2} ( 10 )} \cdot 1 0^{- z} + C

Solution steps

\int \frac{z}{1 0 ^{z}} d z

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{z}{1 0 ^{z}} = (z) 1 0^{- z}

= \int z \cdot 1 0^{- z} d z

Apply Integration By Parts

: - \frac{1 0 ^{- z} z}{ln ( 10 )} - \int - \frac{1 0 ^{- z}}{ln ( 10 )} d z

= - \frac{1 0 ^{- z} z}{ln ( 10 )} - \int - \frac{1 0 ^{- z}}{ln ( 10 )} d z

\int - \frac{1 0 ^{- z}}{ln ( 10 )} d z = \frac{1}{ln ^{2} ( 10 )} \cdot 1 0^{- z}

= - \frac{1 0 ^{- z} z}{ln ( 10 )} - \frac{1}{ln ^{2} ( 10 )} \cdot 1 0^{- z}

Add a constant to the solution

= - \frac{1 0 ^{- z} z}{ln ( 10 )} - \frac{1}{ln ^{2} ( 10 )} \cdot 1 0^{- z} + C

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What is the integral of z/(10^z) ?
The integral of z/(10^z) is -(10^{-z}z)/(ln(10))-1/(ln^2(10))*10^{-z}+C