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

The integral of 1/(xln(x)sqrt(ln^2(x)-4)) is 1/2 arcsec(1/2 ln(x))+C

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integral of 1/(xln(x)sqrt(ln^2(x)-4))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u u ^{2} - 4} d u

= \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} ln (x))

Simplify

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

Add a constant to the solution

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

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What is the integral of 1/(xln(x)sqrt(ln^2(x)-4)) ?
The integral of 1/(xln(x)sqrt(ln^2(x)-4)) is 1/2 arcsec(1/2 ln(x))+C