What is the integral of 1/(x*ln(x^9)) ?

The integral of 1/(x*ln(x^9)) is 1/9 ln(9|ln(x)|)+C

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integral of 1/(x*ln(x^9))

\int \frac{1}{x \cdot ln ( x ^{9} )} d x

\frac{1}{9} ln (9 ∣ ln (x) ∣) + C

Solution steps

\int \frac{1}{x ln ( x ^{9} )} d x

Apply u-substitution

: \int \frac{1}{9 u ln ( u )} d u

= \int \frac{1}{9 u ln ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{9} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{9} ln ∣ v ∣

Substitute back

= \frac{1}{9} ln ln (x^{9})

Simplify

\frac{1}{9} ln ln (x^{9}) : \frac{1}{9} ln (9 ∣ ln (x) ∣)

= \frac{1}{9} ln (9 ∣ ln (x) ∣)

Add a constant to the solution

= \frac{1}{9} ln (9 ∣ ln (x) ∣) + C

y^{^{'}} + 2 x y = x \cdot e^{- x^{2}}

\frac{\partial}{\partial x} (2 ln ((x + y)^{3}))

\frac{d}{d x} ((14 x - x^{2})^{3})

f (x) = 8 x^{2} + 8 x - 5, at x = - \frac{1}{2}

\frac{d}{d x} (\frac{4 x}{5 - tan ( x )})

What is the integral of 1/(x*ln(x^9)) ?
The integral of 1/(x*ln(x^9)) is 1/9 ln(9|ln(x)|)+C