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

\frac{\partial}{\partial y} (x^{3} + y^{2} - y x^{2} - x y)

x \to 1 - lim ((x - 1) ∣ x - 1 ∣)

f (x) = 4 - 6 x

\int_{2}^{2} x^{3} d x

\frac{d y}{d t} = \frac{t e ^{t}}{y 1 + y ^{2}}

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