What is the integral of (sec(x)tan(x))/(a+bsec(x)) ?

The integral of (sec(x)tan(x))/(a+bsec(x)) is 1/b ln|a+bsec(x)|+C

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integral of (sec(x)tan(x))/(a+bsec(x))

\int \frac{sec ( x ) tan ( x )}{a + b sec ( x )} d x

\frac{1}{b} ln ∣ a + b sec (x) ∣ + C

Solution steps

\int \frac{sec ( x ) tan ( x )}{a + b sec ( x )} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} ln ∣ u ∣

Substitute back

u = a + b sec (x)

= \frac{1}{b} ln ∣ a + b sec (x) ∣

Add a constant to the solution

= \frac{1}{b} ln ∣ a + b sec (x) ∣ + C

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What is the integral of (sec(x)tan(x))/(a+bsec(x)) ?
The integral of (sec(x)tan(x))/(a+bsec(x)) is 1/b ln|a+bsec(x)|+C