What is the integral of 1/(x-3x^2) ?

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

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integral of 1/(x-3x^2)

\int \frac{1}{x - 3 x ^{2}} d x

- ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣) + C

Solution steps

\int \frac{1}{x - 3 x ^{2}} d x

Factor out common term

- x : - x (3 x - 1)

= \int \frac{1}{- x ( 3 x - 1 )} d x

Take the constant out:

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

= - \int \frac{1}{x ( 3 x - 1 )} d x

Apply u-substitution

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

Take the partial fraction of

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

= - \int \frac{1}{u} - \frac{1}{u + 1} d u

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}{u} d u - \int \frac{1}{u + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

Substitute back

u = 3 x - 1

= - (ln ∣ 3 x - 1 ∣ - ln ∣ 3 x - 1 + 1 ∣)

Simplify

- (ln ∣ 3 x - 1 ∣ - ln ∣ 3 x - 1 + 1 ∣) : - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣)

= - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣)

Add a constant to the solution

= - ln ∣ 3 x - 1 ∣ + ln (3 ∣ x ∣) + C

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