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

sin (w)

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + e^{(- x)}

y^{^{'}} = 2 x e^{(2 y)}

x \to \frac{π}{2} - lim (sec (x))

y = g (x), at x = 6

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