What is the integral of 1/((x^2-4x)) ?

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

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integral of 1/((x^2-4x))

integral

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

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

Solution steps

\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 )}

= \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

= - \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 ∣

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

Add a constant to the solution

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

in t e g r a l (2 x + 3)^{4}

in t e g r a l e \cdot x^{2}

in t e g r a l e^{2 x^{2}}

in t e g r a l x^{2} + 6 x

in t e g r a l \frac{2 x + 1}{x}

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