What is the integral from 4 to 7 of 1/(4x^2-8x-5) ?

The integral from 4 to 7 of 1/(4x^2-8x-5) is -(ln(5)-2ln(3))/(12)

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integral from 4 to 7 of 1/(4x^2-8x-5)

\int_{4}^{7} \frac{1}{4 x ^{2} - 8 x - 5} d x

- \frac{ln ( 5 ) - 2 ln ( 3 )}{12}

Decimal Notation

0.04898 \dots

Solution steps

\int_{4}^{7} \frac{1}{4 x ^{2} - 8 x - 5} d x

Complete the square

4 x^{2} - 8 x - 5 : 4 (x - 1)^{2} - 9

= \int_{4}^{7} \frac{1}{4 ( x - 1 ) ^{2} - 9} d x

Apply u-substitution

: \int_{3}^{6} \frac{1}{4 u ^{2} - 9} d u

= \int_{3}^{6} \frac{1}{4 u ^{2} - 9} d u

Apply Integral Substitution

= \int_{2}^{4} \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_{2}^{4} \frac{1}{v ^{2} - 1} d v

Factor

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

= \frac{1}{6} \cdot \int_{2}^{4} \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_{2}^{4} \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}]_{2}^{4})

Simplify

\frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{2}^{4}) : - \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{2}^{4}

= - \frac{1}{6} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{2}^{4}

Compute the boundaries:

\frac{ln ( 5 ) - 2 ln ( 3 )}{2}

= - \frac{1}{6} \cdot \frac{ln ( 5 ) - 2 ln ( 3 )}{2}

Simplify

= - \frac{ln ( 5 ) - 2 ln ( 3 )}{12}

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What is the integral from 4 to 7 of 1/(4x^2-8x-5) ?
The integral from 4 to 7 of 1/(4x^2-8x-5) is -(ln(5)-2ln(3))/(12)