What is the integral of (ln(\sqrt[4]{x}))/(x^2) ?

The integral of (ln(\sqrt[4]{x}))/(x^2) is (-ln(x)-1)/(4x)+C

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integral of (ln(\sqrt[4]{x}))/(x^2)

\int \frac{ln ( 4 x )}{x ^{2}} d x

\frac{- ln ( x ) - 1}{4 x} + C

Solution steps

\int \frac{ln ( 4 x )}{x ^{2}} d x

Simplify

\frac{ln ( 4 x )}{x ^{2}} : \frac{ln ( x )}{4 x ^{2}}

= \int \frac{ln ( x )}{4 x ^{2}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x

= \frac{1}{4} (- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x)

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

= \frac{1}{4} (- \frac{ln ( x )}{x} - \frac{1}{x})

Simplify

\frac{1}{4} (- \frac{ln ( x )}{x} - \frac{1}{x}) : \frac{- ln ( x ) - 1}{4 x}

= \frac{- ln ( x ) - 1}{4 x}

Add a constant to the solution

= \frac{- ln ( x ) - 1}{4 x} + C

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What is the integral of (ln(\sqrt[4]{x}))/(x^2) ?
The integral of (ln(\sqrt[4]{x}))/(x^2) is (-ln(x)-1)/(4x)+C