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

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

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

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

- ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

\frac{2 x - 1}{( x - 1 ) ( x - 2 )} : \frac{2 x}{( x - 1 ) ( x - 2 )} - \frac{1}{( x - 1 ) ( x - 2 )}

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

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

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

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

Simplify

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

= - ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + 3 ln ∣ x - 2 ∣ + C

\frac{e ^{x}}{\frac{d y}{d x}} = (1 + e^{x}) y

y^{^{''}} - 6 y^{^{'}} + 25 y = 50 t^{3} - 36 t^{2} - 63 t + 18

\frac{\partial}{\partial x} (ln (t))

x \to + 2 lim (\frac{x - 7}{x + 3})

\frac{2 x - 7}{5 x + 4} + x^{3}

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