What is the integral of (18)/(x[(ln(x))^2-ln(x^9)]) ?

The integral of (18)/(x[(ln(x))^2-ln(x^9)]) is 18(-1/9 ln|ln(x)|+1/9 ln|ln(x)-9|)+C

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integral of (18)/(x[(ln(x))^2-ln(x^9)])

\int \frac{18}{x [ ( ln ( x ) ) ^{2} - ln ( x ^{9} ) ]} d x

18 (- \frac{1}{9} ln ∣ ln (x) ∣ + \frac{1}{9} ln ∣ ln (x) - 9 ∣) + C

Solution steps

\int \frac{18}{x [ ( ln ( x ) ) ^{2} - ln ( x ^{9} ) ]} d x

Take the constant out:

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

= 18 \cdot \int \frac{1}{x [ ( ln ( x ) ) ^{2} - ln ( x ^{9} ) ]} d x

Apply u-substitution

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

= 18 \cdot \int \frac{1}{u ^{2} - 9 u} d u

Take the partial fraction of

\frac{1}{u ^{2} - 9 u} : - \frac{1}{9 u} + \frac{1}{9 ( u - 9 )}

= 18 \cdot \int - \frac{1}{9 u} + \frac{1}{9 ( u - 9 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 18 (- \int \frac{1}{9 u} d u + \int \frac{1}{9 ( u - 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u - 9 )} d u = \frac{1}{9} ln ∣ u - 9 ∣

= 18 (- \frac{1}{9} ln ∣ u ∣ + \frac{1}{9} ln ∣ u - 9 ∣)

Substitute back

u = ln (x)

= 18 (- \frac{1}{9} ln ∣ ln (x) ∣ + \frac{1}{9} ln ∣ ln (x) - 9 ∣)

Add a constant to the solution

= 18 (- \frac{1}{9} ln ∣ ln (x) ∣ + \frac{1}{9} ln ∣ ln (x) - 9 ∣) + C

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What is the integral of (18)/(x[(ln(x))^2-ln(x^9)]) ?
The integral of (18)/(x[(ln(x))^2-ln(x^9)]) is 18(-1/9 ln|ln(x)|+1/9 ln|ln(x)-9|)+C