What is the integral of (2x-4)/(x^2-x) ?

The integral of (2x-4)/(x^2-x) is -2ln|x-1|+4ln|x|+C

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integral of (2x-4)/(x^2-x)

\int \frac{2 x - 4}{x ^{2} - x} d x

- 2 ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{2 x - 4}{x ^{2} - x} d x

Expand

\frac{2 x - 4}{x ^{2} - x} : \frac{2 x}{x ^{2} - x} - \frac{4}{x ^{2} - 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{2 x}{x ^{2} - x} d x - \int \frac{4}{x ^{2} - x} d x

\int \frac{2 x}{x ^{2} - x} d x = 2 ln ∣ x - 1 ∣

\int \frac{4}{x ^{2} - x} d x = 4 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

= 2 ln ∣ x - 1 ∣ - 4 (- ln ∣ x ∣ + ln ∣ x - 1 ∣)

Simplify

2 ln ∣ x - 1 ∣ - 4 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) : - 2 ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

= - 2 ln ∣ x - 1 ∣ + 4 ln ∣ x ∣

Add a constant to the solution

= - 2 ln ∣ x - 1 ∣ + 4 ln ∣ x ∣ + C

y^{^{''}} + 10 y^{^{'}} + 25 y = 5 e^{- (5 t)}

\int_{0}^{π} 8 sin (x) d x

(0, 3), (6, 9)

\int 5 x e^{12 x} d x

y^{^{'}} + y = y^{- \frac{1}{2}}

What is the integral of (2x-4)/(x^2-x) ?
The integral of (2x-4)/(x^2-x) is -2ln|x-1|+4ln|x|+C