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

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

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

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

\frac{1}{3} ln ∣ 1 + tan (3 x) ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + tan (3 x)

= \frac{1}{3} ln ∣ 1 + tan (3 x) ∣

Add a constant to the solution

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

y = x^{2} - 3 x - 5, (2, - 7)

f (x) = 3 x^{2} - 12 x + 1, (0, 1)

\frac{d}{d x} (\frac{x ^{3}}{3} - 2 x^{2} - 5 x)

f (x) = x^{2} - 4, at x = 5

n = 0 \sum \infty 3^{- 2 n}

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