What is the integral of (sec^2(6x))/(3tan(6x)+15) ?

The integral of (sec^2(6x))/(3tan(6x)+15) is 1/18 ln|3tan(6x)+15|+C

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integral of (sec^2(6x))/(3tan(6x)+15)

\int \frac{sec ^{2} ( 6 x )}{3 tan ( 6 x ) + 15} d x

\frac{1}{18} ln ∣ 3 tan (6 x) + 15 ∣ + C

Solution steps

\int \frac{sec ^{2} ( 6 x )}{3 tan ( 6 x ) + 15} d x

Apply u-substitution

= \int \frac{1}{18 u} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} ln ∣ u ∣

Substitute back

u = 3 tan (6 x) + 15

= \frac{1}{18} ln ∣ 3 tan (6 x) + 15 ∣

Add a constant to the solution

= \frac{1}{18} ln ∣ 3 tan (6 x) + 15 ∣ + C

What is the integral of (sec^2(6x))/(3tan(6x)+15) ?
The integral of (sec^2(6x))/(3tan(6x)+15) is 1/18 ln|3tan(6x)+15|+C