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

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

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

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

arctan (ln^{2} (x) - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (v)

Substitute back

= arctan (ln^{2} (x) - 1)

Add a constant to the solution

= arctan (ln^{2} (x) - 1) + C

\frac{d y}{d x} - 7 y = e^{x}

\int \frac{x}{x ^{2} - 2 x + 2} d x

x \to 5 + lim (\frac{3 x}{4} + 1)

\frac{d}{d x} (ln (x - a))

\frac{\partial}{\partial x} (\frac{( u - v )}{u + v})

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