What is the integral of log_{10}(1/x) ?

The integral of log_{10}(1/x) is -xlog_{10}(x)+1/(ln(10))x+C

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integral of log_{10}(1/x)

integral

lo g_{10} (\frac{1}{x})

- x lo g_{10} (x) + \frac{1}{ln ( 10 )} x + C

Solution steps

\int lo g_{10} (\frac{1}{x}) d x

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= \int - lo g_{10} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int lo g_{10} (x) d x

Apply Integration By Parts

: x lo g_{10} (x) - \int \frac{1}{ln ( 10 )} d x

= - (x lo g_{10} (x) - \int \frac{1}{ln ( 10 )} d x)

\int \frac{1}{ln ( 10 )} d x = \frac{1}{ln ( 10 )} x

= - (x lo g_{10} (x) - \frac{1}{ln ( 10 )} x)

Simplify

= - x lo g_{10} (x) + \frac{1}{ln ( 10 )} x

Add a constant to the solution

= - x lo g_{10} (x) + \frac{1}{ln ( 10 )} x + C

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What is the integral of log_{10}(1/x) ?
The integral of log_{10}(1/x) is -xlog_{10}(x)+1/(ln(10))x+C