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

The integral of ((x+1))/((x-1)) is 2ln|x-1|+x+C

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

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

2 ln ∣ x - 1 ∣ + x + C

Solution steps

\int \frac{x + 1}{x - 1} d x

\frac{x + 1}{x - 1} = \frac{2}{x - 1} + 1

= \int \frac{2}{x - 1} + 1 d x

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{2}{x - 1} d x + \int 1 d x

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

\int 1 d x = x

= 2 ln ∣ x - 1 ∣ + x

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ + x + C

x \to 1 + lim (2)

f (x) = x, (16, 4)

x \to 5 lim (2 x^{2} - 5 x + 3)

y = x + 8

\frac{d}{d x} (\frac{- 4 x}{x ^{2} + 1})

What is the integral of ((x+1))/((x-1)) ?
The integral of ((x+1))/((x-1)) is 2ln|x-1|+x+C