What is the integral of ((2x+1))/((x-1)) ?

The integral of ((2x+1))/((x-1)) is 2x-2+3ln|x-1|+C

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integral of ((2x+1))/((x-1))

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

2 x - 2 + 3 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

= 2 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣

Simplify

2 (x - 1 + ln ∣ x - 1 ∣) + ln ∣ x - 1 ∣ : 2 x - 2 + 3 ln ∣ x - 1 ∣

= 2 x - 2 + 3 ln ∣ x - 1 ∣

Add a constant to the solution

= 2 x - 2 + 3 ln ∣ x - 1 ∣ + C

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\int (y - 1)^{- 2} d y

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What is the integral of ((2x+1))/((x-1)) ?
The integral of ((2x+1))/((x-1)) is 2x-2+3ln|x-1|+C