What is the integral of (sec(x)tan(x))/(sec(x)-4) ?

The integral of (sec(x)tan(x))/(sec(x)-4) is ln|sec(x)-4|+C

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integral of (sec(x)tan(x))/(sec(x)-4)

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 4} d x

ln ∣ sec (x) - 4 ∣ + C

Solution steps

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 4} 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) - 4

= ln ∣ sec (x) - 4 ∣

Add a constant to the solution

= ln ∣ sec (x) - 4 ∣ + C

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 4

\int \frac{1}{( 1 - t ) ^{2}} d t

\frac{d}{d x} (3 sin (6 x))

\int \frac{1}{2} x^{3} 4 - x^{2} d x

y = \frac{1}{x ^{8}}

What is the integral of (sec(x)tan(x))/(sec(x)-4) ?
The integral of (sec(x)tan(x))/(sec(x)-4) is ln|sec(x)-4|+C