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

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

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

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

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

Solution steps

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

Factor

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

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

Take the constant out:

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

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

Simplify

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

Take the partial fraction of

\frac{1}{x ( x - 1 )} : - \frac{1}{x} + \frac{1}{x - 1}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

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