What is the integral of (3sec^2(x))/(9tan(x)-1) ?

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

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integral of (3sec^2(x))/(9tan(x)-1)

\int \frac{3 sec ^{2} ( x )}{9 tan ( x ) - 1} d x

\frac{1}{3} ln ∣ 9 tan (x) - 1 ∣ + C

Solution steps

\int \frac{3 sec ^{2} ( x )}{9 tan ( x ) - 1} d x

Take the constant out:

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

= 3 \cdot \int \frac{sec ^{2} ( x )}{9 tan ( x ) - 1} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{9 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 tan (x) - 1

= 3 \cdot \frac{1}{9} ln ∣ 9 tan (x) - 1 ∣

Simplify

3 \cdot \frac{1}{9} ln ∣ 9 tan (x) - 1 ∣ : \frac{1}{3} ln ∣ 9 tan (x) - 1 ∣

= \frac{1}{3} ln ∣ 9 tan (x) - 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 9 tan (x) - 1 ∣ + C

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