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

The integral of 1/(x(ln(x^2))) is 1/2 ln(2|ln(x)|)+C

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

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

\frac{1}{2} ln (2 ∣ ln (x) ∣) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ln ( u )} d u

= \int \frac{1}{2 u ln ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

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

= \frac{1}{2} ln (2 ∣ ln (x) ∣)

Add a constant to the solution

= \frac{1}{2} ln (2 ∣ ln (x) ∣) + C

\int ln (e^{x y} cos (x)) d y

\int x^{4} a x^{5} - b d x

\frac{\partial}{\partial x} (ln (x + 6 y + 9 z))

\int_{0}^{\frac{π}{2}} tan (\frac{x}{2}) d x

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

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