What is the integral of 1/(4x^2-14x-1) ?

The integral of 1/(4x^2-14x-1) is -1/(2sqrt(53))(ln|(4x-7)/(sqrt(53))+1|-ln|(4x-7)/(sqrt(53))-1|)+C

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integral of 1/(4x^2-14x-1)

\int \frac{1}{4 x ^{2} - 14 x - 1} d x

- \frac{1}{2 53} (ln \frac{4 x - 7}{53} + 1 - ln \frac{4 x - 7}{53} - 1) + C

Solution steps

\int \frac{1}{4 x ^{2} - 14 x - 1} d x

Factor

4 x^{2} - 14 x - 1 : - (- 4 x^{2} + 14 x + 1)

= \int \frac{1}{- ( - 4 x ^{2} + 14 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}{- 4 x ^{2} + 14 x + 1} d x

Complete the square

- 4 x^{2} + 14 x + 1 : - 4 (x - \frac{7}{4})^{2} + \frac{53}{4}

= - \int \frac{1}{- 4 ( x - \frac{7}{4} ) ^{2} + \frac{53}{4}} d x

Apply u-substitution

: \int \frac{4}{- 16 u ^{2} + 53} d u

= - \int \frac{4}{- 16 u ^{2} + 53} 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}{- 16 u ^{2} + 53} d u

Apply Integral Substitution

= - 4 \cdot \int \frac{1}{4 53 ( - 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}{4 53} \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}{4 53} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - 4 \cdot \frac{1}{4 53} \frac{ln \frac{4}{53} ( x - \frac{7}{4} ) + 1}{2} - \frac{ln \frac{4}{53} ( x - \frac{7}{4} ) - 1}{2}

Simplify

- 4 \cdot \frac{1}{4 53} \frac{ln \frac{4}{53} ( x - \frac{7}{4} ) + 1}{2} - \frac{ln \frac{4}{53} ( x - \frac{7}{4} ) - 1}{2} : - \frac{1}{2 53} (ln \frac{4 x - 7}{53} + 1 - ln \frac{4 x - 7}{53} - 1)

= - \frac{1}{2 53} (ln \frac{4 x - 7}{53} + 1 - ln \frac{4 x - 7}{53} - 1)

Add a constant to the solution

= - \frac{1}{2 53} (ln \frac{4 x - 7}{53} + 1 - ln \frac{4 x - 7}{53} - 1) + C

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What is the integral of 1/(4x^2-14x-1) ?
The integral of 1/(4x^2-14x-1) is -1/(2sqrt(53))(ln|(4x-7)/(sqrt(53))+1|-ln|(4x-7)/(sqrt(53))-1|)+C