What is the integral of log_{3}(\sqrt[5]{2x}) ?

The integral of log_{3}(\sqrt[5]{2x}) is 1/5 (xlog_{3}(2x)-1/(ln(3))x)+C

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integral of log_{3}(\sqrt[5]{2x})

\frac{1}{5} (x lo g_{3} (2 x) - \frac{1}{ln ( 3 )} x) + C

Solution steps

Simplify

= \int \frac{1}{5} lo g_{3} (2 x) d x

Take the constant out:

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

= \frac{1}{5} \cdot \int lo g_{3} (2 x) d x

Apply Integration By Parts

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

= \frac{1}{5} (x lo g_{3} (2 x) - \int \frac{1}{ln ( 3 )} d x)

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

= \frac{1}{5} (x lo g_{3} (2 x) - \frac{1}{ln ( 3 )} x)

Add a constant to the solution

= \frac{1}{5} (x lo g_{3} (2 x) - \frac{1}{ln ( 3 )} x) + C

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What is the integral of log_{3}(\sqrt[5]{2x}) ?
The integral of log_{3}(\sqrt[5]{2x}) is 1/5 (xlog_{3}(2x)-1/(ln(3))x)+C