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

The integral of ((x+2)/(x^2-4x)) is -1/2 ln|x|+3/2 ln|x-4|-1+C

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

\int (\frac{x + 2}{x ^{2} - 4 x}) d x

- \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 4 ∣ - 1 + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

(x + 2) (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣) - \frac{1}{4} (- x ln ∣ x ∣ + x ln ∣ x - 4 ∣ - 4 ln ∣ x - 4 ∣ + 4) : - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 4 ∣ - 1

= - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 4 ∣ - 1

Add a constant to the solution

= - \frac{1}{2} ln ∣ x ∣ + \frac{3}{2} ln ∣ x - 4 ∣ - 1 + C

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What is the integral of ((x+2)/(x^2-4x)) ?
The integral of ((x+2)/(x^2-4x)) is -1/2 ln|x|+3/2 ln|x-4|-1+C