What is the integral of 1/(x(ln(x^2))^6) ?

The integral of 1/(x(ln(x^2))^6) is -1/(320ln^5(x))+C

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integral of 1/(x(ln(x^2))^6)

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{6}} d x

- \frac{1}{320 ln ^{5} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{6}} d x

Simplify

\frac{1}{x ( ln ( x ^{2} ) ) ^{6}} : \frac{1}{64 x ln ^{6} ( x )}

= \int \frac{1}{64 x ln ^{6} ( x )} d x

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{x ln ^{6} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{6}} d u

= \frac{1}{64} \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= \frac{1}{64} (- \frac{1}{5 u ^{5}})

Substitute back

u = ln (x)

= \frac{1}{64} (- \frac{1}{5 ln ^{5} ( x )})

Simplify

\frac{1}{64} (- \frac{1}{5 ln ^{5} ( x )}) : - \frac{1}{320 ln ^{5} ( x )}

= - \frac{1}{320 ln ^{5} ( x )}

Add a constant to the solution

= - \frac{1}{320 ln ^{5} ( x )} + C

\int \frac{5 x ^{2} + 1}{x - 1} d x

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\int \frac{y ^{3} - 4 y + 3}{y + 5} d y

What is the integral of 1/(x(ln(x^2))^6) ?
The integral of 1/(x(ln(x^2))^6) is -1/(320ln^5(x))+C