What is the integral of 1/(x(ln(x^2))^3) ?

The integral of 1/(x(ln(x^2))^3) is -1/(16ln^2(x))+C

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integral of 1/(x(ln(x^2))^3)

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{3}} d x

- \frac{1}{16 ln ^{2} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{3}} d x

Simplify

\frac{1}{x ( ln ( x ^{2} ) ) ^{3}} : \frac{1}{8 x ln ^{3} ( x )}

= \int \frac{1}{8 x ln ^{3} ( x )} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{x ln ^{3} ( x )} d x

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \frac{1}{8} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{8} (- \frac{1}{2 u ^{2}})

Substitute back

u = ln (x)

= \frac{1}{8} (- \frac{1}{2 ln ^{2} ( x )})

Simplify

\frac{1}{8} (- \frac{1}{2 ln ^{2} ( x )}) : - \frac{1}{16 ln ^{2} ( x )}

= - \frac{1}{16 ln ^{2} ( x )}

Add a constant to the solution

= - \frac{1}{16 ln ^{2} ( x )} + C

x \to 5 + lim (x^{2} - 25)

\int 0.6 x^{2} d x

x \to + 0 lim (ln (x))

\frac{3 x}{1 - x}

\int sec (x) - 1 d x

What is the integral of 1/(x(ln(x^2))^3) ?
The integral of 1/(x(ln(x^2))^3) is -1/(16ln^2(x))+C