What is the integral of 1/(6x^2-5x-6) ?

The integral of 1/(6x^2-5x-6) is -1/13 (ln| 8/13+12/13 x|-ln|-18/13+12/13 x|)+C

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integral of 1/(6x^2-5x-6)

\int \frac{1}{6 x ^{2} - 5 x - 6} d x

- \frac{1}{13} (ln \frac{8}{13} + \frac{12}{13} x - ln - \frac{18}{13} + \frac{12}{13} x) + C

Solution steps

\int \frac{1}{6 x ^{2} - 5 x - 6} d x

Complete the square

6 x^{2} - 5 x - 6 : 6 (x - \frac{5}{12})^{2} - \frac{169}{24}

= \int \frac{1}{6 ( x - \frac{5}{12} ) ^{2} - \frac{169}{24}} d x

Apply u-substitution

= \int \frac{24}{144 u ^{2} - 169} d u

Take the constant out:

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

= 24 \cdot \int \frac{1}{144 u ^{2} - 169} d u

Apply Integral Substitution

= 24 \cdot \int \frac{1}{156 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 24 \cdot \frac{1}{156} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 24 \cdot \frac{1}{156} \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

= 24 \cdot \frac{1}{156} (- \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}

= 24 \cdot \frac{1}{156} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 24 \cdot \frac{1}{156} (- (\frac{ln \frac{12}{13} ( x - \frac{5}{12} ) + 1}{2} - \frac{ln \frac{12}{13} ( x - \frac{5}{12} ) - 1}{2}))

Simplify

24 \cdot \frac{1}{156} (- (\frac{ln \frac{12}{13} ( x - \frac{5}{12} ) + 1}{2} - \frac{ln \frac{12}{13} ( x - \frac{5}{12} ) - 1}{2})) : - \frac{1}{13} (ln \frac{8}{13} + \frac{12}{13} x - ln - \frac{18}{13} + \frac{12}{13} x)

= - \frac{1}{13} (ln \frac{8}{13} + \frac{12}{13} x - ln - \frac{18}{13} + \frac{12}{13} x)

Add a constant to the solution

= - \frac{1}{13} (ln \frac{8}{13} + \frac{12}{13} x - ln - \frac{18}{13} + \frac{12}{13} x) + C

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What is the integral of 1/(6x^2-5x-6) ?
The integral of 1/(6x^2-5x-6) is -1/13 (ln| 8/13+12/13 x|-ln|-18/13+12/13 x|)+C