What is the integral of (tan(x)+1)/(tan(x)-1) ?

The integral of (tan(x)+1)/(tan(x)-1) is ln|sin(x)-cos(x)|+C

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integral of (tan(x)+1)/(tan(x)-1)

\int \frac{tan ( x ) + 1}{tan ( x ) - 1} d x

ln ∣ sin (x) - cos (x) ∣ + C

Solution steps

\int \frac{tan ( x ) + 1}{tan ( x ) - 1} d x

Rewrite using trig identities

= \int \frac{sin ( x ) + cos ( x )}{sin ( x ) - cos ( x )} 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 = sin (x) - cos (x)

= ln ∣ sin (x) - cos (x) ∣

Add a constant to the solution

= ln ∣ sin (x) - cos (x) ∣ + C

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What is the integral of (tan(x)+1)/(tan(x)-1) ?
The integral of (tan(x)+1)/(tan(x)-1) is ln|sin(x)-cos(x)|+C