What is the integral of 1/(x(ln(x^2))^9) ?

The integral of 1/(x(ln(x^2))^9) is -1/(4096ln^8(x))+C

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integral of 1/(x(ln(x^2))^9)

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{9}} d x

- \frac{1}{4096 ln ^{8} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{9}} d x

Simplify

\frac{1}{x ( ln ( x ^{2} ) ) ^{9}} : \frac{1}{512 x ln ^{9} ( x )}

= \int \frac{1}{512 x ln ^{9} ( x )} d x

Take the constant out:

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

= \frac{1}{512} \cdot \int \frac{1}{x ln ^{9} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{9}} d u

= \frac{1}{512} \cdot \int \frac{1}{u ^{9}} d u

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= \frac{1}{512} (- \frac{1}{8 u ^{8}})

Substitute back

u = ln (x)

= \frac{1}{512} (- \frac{1}{8 ln ^{8} ( x )})

Simplify

\frac{1}{512} (- \frac{1}{8 ln ^{8} ( x )}) : - \frac{1}{4096 ln ^{8} ( x )}

= - \frac{1}{4096 ln ^{8} ( x )}

Add a constant to the solution

= - \frac{1}{4096 ln ^{8} ( x )} + C

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What is the integral of 1/(x(ln(x^2))^9) ?
The integral of 1/(x(ln(x^2))^9) is -1/(4096ln^8(x))+C