What is the integral of 1/(3x^2-5x+1) ?

The integral of 1/(3x^2-5x+1) is -1/(sqrt(13))(ln|(6x-5)/(sqrt(13))+1|-ln|(6x-5)/(sqrt(13))-1|)+C

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integral of 1/(3x^2-5x+1)

\int \frac{1}{3 x ^{2} - 5 x + 1} d x

- \frac{1}{13} (ln \frac{6 x - 5}{13} + 1 - ln \frac{6 x - 5}{13} - 1) + C

Solution steps

\int \frac{1}{3 x ^{2} - 5 x + 1} d x

Complete the square

3 x^{2} - 5 x + 1 : 3 (x - \frac{5}{6})^{2} - \frac{13}{12}

= \int \frac{1}{3 ( x - \frac{5}{6} ) ^{2} - \frac{13}{12}} d x

Apply u-substitution

: \int \frac{12}{36 u ^{2} - 13} d u

= \int \frac{12}{36 u ^{2} - 13} d u

Take the constant out:

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

= 12 \cdot \int \frac{1}{36 u ^{2} - 13} d u

Apply Integral Substitution

= 12 \cdot \int \frac{1}{6 13 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 12 \cdot \frac{1}{6 13} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= 12 \cdot \frac{1}{6 13} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 12 \cdot \frac{1}{6 13} (- \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}

= 12 \cdot \frac{1}{6 13} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 12 \cdot \frac{1}{6 13} - \frac{ln \frac{6}{13} ( x - \frac{5}{6} ) + 1}{2} - \frac{ln \frac{6}{13} ( x - \frac{5}{6} ) - 1}{2}

Simplify

12 \cdot \frac{1}{6 13} - \frac{ln \frac{6}{13} ( x - \frac{5}{6} ) + 1}{2} - \frac{ln \frac{6}{13} ( x - \frac{5}{6} ) - 1}{2} : - \frac{1}{13} (ln \frac{6 x - 5}{13} + 1 - ln \frac{6 x - 5}{13} - 1)

= - \frac{1}{13} (ln \frac{6 x - 5}{13} + 1 - ln \frac{6 x - 5}{13} - 1)

Add a constant to the solution

= - \frac{1}{13} (ln \frac{6 x - 5}{13} + 1 - ln \frac{6 x - 5}{13} - 1) + C

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What is the integral of 1/(3x^2-5x+1) ?
The integral of 1/(3x^2-5x+1) is -1/(sqrt(13))(ln|(6x-5)/(sqrt(13))+1|-ln|(6x-5)/(sqrt(13))-1|)+C