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

\frac{d ^{3}}{d x ^{3}} (\frac{8}{x ^{2}})

\int 9 x e^{7 x} d x

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 7 e^{x}

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

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