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

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

f (x) = arcsin (2 x)

\int_{0}^{1} 7 e^{- 4 x} d x

x \to 3 lim (lo g_{4} (3 x - 5))

x \to - 5 + lim (\frac{1}{x + 5})

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