What is the integral of log_{6}(x) ?

The integral of log_{6}(x) is xlog_{6}(x)-1/(ln(6))x+C

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integral of log_{6}(x)

\int lo g_{6} (x) d x

x lo g_{6} (x) - \frac{1}{ln ( 6 )} x + C

Solution steps

\int lo g_{6} (x) d x

Apply Integration By Parts

: x lo g_{6} (x) - \int \frac{1}{ln ( 6 )} d x

= x lo g_{6} (x) - \int \frac{1}{ln ( 6 )} d x

\int \frac{1}{ln ( 6 )} d x = \frac{1}{ln ( 6 )} x

= x lo g_{6} (x) - \frac{1}{ln ( 6 )} x

Add a constant to the solution

= x lo g_{6} (x) - \frac{1}{ln ( 6 )} x + C

\frac{\partial}{\partial x} (4 x y e^{- x^{2} - y^{2}})

\frac{\partial}{\partial x} (x y + 2 x - ln (x^{2} y))

\int (x^{2} + 3) d x

(x^{4} + 1)^{5}

\frac{d}{d x} (\frac{5}{( 2 x ) ^{3}} + 2 cos (x))

What is the integral of log_{6}(x) ?
The integral of log_{6}(x) is xlog_{6}(x)-1/(ln(6))x+C