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

n = 0 \sum \infty \frac{1}{n ^{2} + n}

\int \frac{x}{x ^{2} - 6 x} d x

y^{^{'}} + y = \frac{1}{1 + e ^{t}}

y^{^{'}} + 7 x^{6} y = x^{6}

x \to 0 lim (x^{2})

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