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

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

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

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

2 arctan (2 ln (x) - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 arctan (v)

Substitute back

= 2 arctan (2 ln (x) - 1)

Add a constant to the solution

= 2 arctan (2 ln (x) - 1) + C

sin^{2} (θ)

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\int x 23 x - 16 d x

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