Question

How do you find the indefinite integral of `int (secxtanx)/(secx-1)`?

Answer 1

`int(secxtanx)/(secx-1)dx=ln|(secx-1)|+C`

Explanation:

we use the standard result

`int(f'(x))/(f(x))dx=ln|f(x)|+C`

`int(secxtanx)/(secx-1)dx`

differentiate the denominataor:`" "d/(dx)(secx-1)=secxtanx`

`:.int(secxtanx)/(secx-1)dx=ln|(secx-1)|+C`