What is the integral of log_{8}(3x) ?

The integral of log_{8}(3x) is xlog_{8}(3x)-1/(3ln(2))x+C

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integral of log_{8}(3x)

\int lo g_{8} (3 x) d x

x lo g_{8} (3 x) - \frac{1}{3 ln ( 2 )} x + C

Solution steps

\int lo g_{8} (3 x) d x

Apply Integration By Parts

: x lo g_{8} (3 x) - \int \frac{1}{3 ln ( 2 )} d x

= x lo g_{8} (3 x) - \int \frac{1}{3 ln ( 2 )} d x

\int \frac{1}{3 ln ( 2 )} d x = \frac{1}{3 ln ( 2 )} x

= x lo g_{8} (3 x) - \frac{1}{3 ln ( 2 )} x

Add a constant to the solution

= x lo g_{8} (3 x) - \frac{1}{3 ln ( 2 )} x + C

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What is the integral of log_{8}(3x) ?
The integral of log_{8}(3x) is xlog_{8}(3x)-1/(3ln(2))x+C