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

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

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

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

\frac{1}{3} (ln (\frac{2}{3} ∣ x ∣) - ln \frac{2}{3} x - 2) + C

Solution steps

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

Complete the square

3 x - x^{2} : - (x - \frac{3}{2})^{2} + \frac{9}{4}

= \int \frac{1}{- ( x - \frac{3}{2} ) ^{2} + \frac{9}{4}} d x

Apply u-substitution

: \int \frac{4}{- 4 u ^{2} + 9} d u

= \int \frac{4}{- 4 u ^{2} + 9} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{- 4 u ^{2} + 9} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{6 ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 4 \cdot \frac{1}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= 4 \cdot \frac{1}{6} (\frac{ln \frac{2}{3} ( x - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( x - \frac{3}{2} ) - 1}{2})

Simplify

4 \cdot \frac{1}{6} (\frac{ln \frac{2}{3} ( x - \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{3} ( x - \frac{3}{2} ) - 1}{2}) : \frac{1}{3} (ln (\frac{2}{3} ∣ x ∣) - ln \frac{2}{3} x - 2)

= \frac{1}{3} (ln (\frac{2}{3} ∣ x ∣) - ln \frac{2}{3} x - 2)

Add a constant to the solution

= \frac{1}{3} (ln (\frac{2}{3} ∣ x ∣) - ln \frac{2}{3} x - 2) + C

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