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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{2} - 4 x} d x

Take the partial fraction of

\frac{1}{x ^{2} - 4 x} : - \frac{1}{4 x} + \frac{1}{4 ( x - 4 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

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

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

Add a constant to the solution

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

sin (2 t) \cdot e^{3 t}

\frac{d}{d x} (arcsin (5 x^{4}))

\int ln (2 + x) d x

(- 1, 1), (1, - 2)

x \to 0 lim (\frac{tan ( 2 ) ( x )}{x})

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