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

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

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

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

- ln ∣ x - 1 ∣ + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= - ln ∣ x - 1 ∣ + x

Add a constant to the solution

= - ln ∣ x - 1 ∣ + x + C

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\frac{d}{d x} (a (2 e^{- x} x - e^{- x} x^{2}))

x \to 1 lim (9 x - 5)

\frac{d}{d x} (x^{3} + x^{2} - 6 x)

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