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

The integral of 1/(2x^2-4x-1) is -1/(2sqrt(6))(ln|sqrt(2/3)(x-1)+1|-ln|sqrt(2/3)(x-1)-1|)+C

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

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

- \frac{1}{2 6} (ln \frac{2}{3} (x - 1) + 1 - ln \frac{2}{3} (x - 1) - 1) + C

Solution steps

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

Complete the square

2 x^{2} - 4 x - 1 : 2 (x - 1)^{2} - 3

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

Apply u-substitution

: \int \frac{1}{2 u ^{2} - 3} d u

= \int \frac{1}{2 u ^{2} - 3} d u

Apply Integral Substitution

= \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

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

Factor

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

= \frac{1}{6} \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

= \frac{1}{6} (- \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}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(2x^2-4x-1) ?
The integral of 1/(2x^2-4x-1) is -1/(2sqrt(6))(ln|sqrt(2/3)(x-1)+1|-ln|sqrt(2/3)(x-1)-1|)+C