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

The integral of (2x+4)/(x^2-x) is 6ln|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

6 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 ∣) : 6 ln ∣ x - 1 ∣ - 4 ln ∣ x ∣

= 6 ln ∣ x - 1 ∣ - 4 ln ∣ x ∣

Add a constant to the solution

= 6 ln ∣ x - 1 ∣ - 4 ln ∣ x ∣ + C

\frac{\partial}{\partial x} (\frac{1}{3} x^{\frac{3}{2}})

\frac{\partial}{\partial y} (6 x cos (x) cos (y))

y^{^{'}} + y = - x

x \to 0 lim \frac{( ( 6 + x ) ^{- 1} - 6 ^{- 1} )}{x}

f (x) = \frac{x ^{2} - 6 x + 12}{x - 4}

What is the integral of (2x+4)/(x^2-x) ?
The integral of (2x+4)/(x^2-x) is 6ln|x-1|-4ln|x|+C