What is the integral of 1/(x(ln(x))^2-9x) ?

The integral of 1/(x(ln(x))^2-9x) is -1/6 (ln|(ln(x))/3+1|-ln|(ln(x))/3-1|)+C

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integral of 1/(x(ln(x))^2-9x)

\int \frac{1}{x ( ln ( x ) ) ^{2} - 9 x} d x

- \frac{1}{6} (ln \frac{ln ( x )}{3} + 1 - ln \frac{ln ( x )}{3} - 1) + C

Solution steps

\int \frac{1}{x ( ln ( x ) ) ^{2} - 9 x} d x

Simplify

= \int \frac{1}{x ln ^{2} ( x ) - 9 x} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 9} d u

= \int \frac{1}{u ^{2} - 9} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} - 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{3} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

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

= \frac{1}{3} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= \frac{1}{3} - \frac{ln \frac{l n ( x )}{3} + 1}{2} - \frac{ln \frac{l n ( x )}{3} - 1}{2}

Simplify

\frac{1}{3} - \frac{ln \frac{l n ( x )}{3} + 1}{2} - \frac{ln \frac{l n ( x )}{3} - 1}{2} : - \frac{1}{6} (ln \frac{ln ( x )}{3} + 1 - ln \frac{ln ( x )}{3} - 1)

= - \frac{1}{6} (ln \frac{ln ( x )}{3} + 1 - ln \frac{ln ( x )}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{ln ( x )}{3} + 1 - ln \frac{ln ( x )}{3} - 1) + C

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What is the integral of 1/(x(ln(x))^2-9x) ?
The integral of 1/(x(ln(x))^2-9x) is -1/6 (ln|(ln(x))/3+1|-ln|(ln(x))/3-1|)+C