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

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

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

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

\frac{1}{2} Ei (ln (x)) + C

Solution steps

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

Simplify

ln (x^{2}) : 2 ln (x)

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

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}{ln ( x )} d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{e ^{u}}{u} d u

Use the nonelementary integral:

\int \frac{e ^{x}}{x} d x = Ei (x)

= \frac{1}{2} Ei (u)

Substitute back

u = ln (x)

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

Add a constant to the solution

= \frac{1}{2} Ei (ln (x)) + C

n = 1 \sum \infty \frac{3 ^{n}}{5 ^{n}}

x \to 0 lim (\frac{9 + x}{x ^{3}})

\frac{\partial}{\partial x} ((2 x^{2} - 5 x)^{3} (2 x + 1))

\int_{0}^{27} 3 x d x

s \to \infty lim (c t)

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