What is the integral of (sec(x)tan(x))/(sec(x)-3) ?

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

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integral of (sec(x)tan(x))/(sec(x)-3)

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 3} d x

ln ∣ sec (x) - 3 ∣ + C

Solution steps

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 3} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = sec (x) - 3

= ln ∣ sec (x) - 3 ∣

Add a constant to the solution

= ln ∣ sec (x) - 3 ∣ + C

x^{2} - 4, [- 2, 0]

\frac{1}{2 - x}

\int cos (x) sin (3 x) d x

\frac{\partial}{\partial y} (x \cdot e^{y})

\frac{d}{d x} (e^{x} - 1)

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