What is the integral of 1/(x(ln(x^2))^4) ?

The integral of 1/(x(ln(x^2))^4) is -1/(48ln^3(x))+C

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integral of 1/(x(ln(x^2))^4)

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{4}} d x

- \frac{1}{48 ln ^{3} ( x )} + C

Solution steps

\int \frac{1}{x ( ln ( x ^{2} ) ) ^{4}} d x

Simplify

\frac{1}{x ( ln ( x ^{2} ) ) ^{4}} : \frac{1}{16 x ln ^{4} ( x )}

= \int \frac{1}{16 x ln ^{4} ( x )} d x

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{x ln ^{4} ( x )} d x

Apply u-substitution

= \frac{1}{16} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

= \frac{1}{16} (- \frac{1}{3 u ^{3}})

Substitute back

u = ln (x)

= \frac{1}{16} (- \frac{1}{3 ln ^{3} ( x )})

Simplify

\frac{1}{16} (- \frac{1}{3 ln ^{3} ( x )}) : - \frac{1}{48 ln ^{3} ( x )}

= - \frac{1}{48 ln ^{3} ( x )}

Add a constant to the solution

= - \frac{1}{48 ln ^{3} ( x )} + C

What is the integral of 1/(x(ln(x^2))^4) ?
The integral of 1/(x(ln(x^2))^4) is -1/(48ln^3(x))+C