What is the integral of 1/(x(ln(x^2))^5) ?

The integral of 1/(x(ln(x^2))^5) is -1/(128ln^4(x))+C

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integral of 1/(x(ln(x^2))^5)

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{5}} d x

- \frac{1}{128 ln ^{4} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{5}} d x

Simplify

\frac{1}{x ( ln ( x ^{2} ) ) ^{5}} : \frac{1}{32 x ln ^{5} ( x )}

= \int \frac{1}{32 x ln ^{5} ( x )} d x

Take the constant out:

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

= \frac{1}{32} \cdot \int \frac{1}{x ln ^{5} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{5}} d u

= \frac{1}{32} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= \frac{1}{32} (- \frac{1}{4 u ^{4}})

Substitute back

u = ln (x)

= \frac{1}{32} (- \frac{1}{4 ln ^{4} ( x )})

Simplify

\frac{1}{32} (- \frac{1}{4 ln ^{4} ( x )}) : - \frac{1}{128 ln ^{4} ( x )}

= - \frac{1}{128 ln ^{4} ( x )}

Add a constant to the solution

= - \frac{1}{128 ln ^{4} ( x )} + C

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What is the integral of 1/(x(ln(x^2))^5) ?
The integral of 1/(x(ln(x^2))^5) is -1/(128ln^4(x))+C