What is the integral of (sec(θ)tan(θ))/(sec^2(θ)-sec(θ)) ?

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

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integral of (sec(θ)tan(θ))/(sec^2(θ)-sec(θ))

\int \frac{sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

- ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣ + C

Solution steps

\int \frac{sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

Apply u-substitution

: \int \frac{1}{u ^{2} - u} d u

= \int \frac{1}{u ^{2} - u} d u

Take the partial fraction of

\frac{1}{u ^{2} - u} : - \frac{1}{u} + \frac{1}{u - 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

Substitute back

u = sec (θ)

= - ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣

Add a constant to the solution

= - ln ∣ sec (θ) ∣ + ln ∣ sec (θ) - 1 ∣ + C

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\frac{\partial}{\partial x} (\frac{( 2 x - y )}{5 x + 3 y})

x^{2} y^{^{''}} + x y^{^{'}} - y = ln (x)

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

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