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

\int x^{3} e^{\frac{x}{2}} d x

\frac{d}{d x} ((3 x^{2} + 7)^{3})

x \to + 2 - lim (x (x - 1))

\int \frac{2 x ^{3}}{( x ^{2} + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{( x + 1 )}{( x ^{3} + x - 2 )})

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