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

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

= 2 x - 2 + ln ∣ x - 1 ∣

Add a constant to the solution

= 2 x - 2 + ln ∣ x - 1 ∣ + C

\int \frac{1}{t ^{2} 9 - t ^{2}} d t

\frac{d ^{2}}{d y ^{2}} (cos (x y^{2}))

\int 5 x ln (3 x) d x

f (x) = arctan (\frac{1}{x ^{2}})

f (x) = x (2 - x) (x + 3)

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