What is the integral from 4 to 5 of 1/(3x^2-3) ?

The integral from 4 to 5 of 1/(3x^2-3) is -(ln(3/2)-ln(5)+ln(3))/6

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integral from 4 to 5 of 1/(3x^2-3)

\int_{4}^{5} \frac{1}{3 x ^{2} - 3} d x

- \frac{ln ( \frac{3}{2} ) - ln ( 5 ) + ln ( 3 )}{6}

Decimal

0.01756 \dots

Solution steps

\int_{4}^{5} \frac{1}{3 x ^{2} - 3} d x

\frac{1}{3 x ^{2} - 3} = \frac{1}{3} \frac{1}{x ^{2} - 1}

= \int_{4}^{5} \frac{1}{3} \cdot \frac{1}{x ^{2} - 1} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{5} \frac{1}{x ^{2} - 1} d x

Factor

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

= \frac{1}{3} \cdot \int_{4}^{5} \frac{1}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= \frac{1}{3} (- \int_{4}^{5} \frac{1}{- x ^{2} + 1} d x)

Use the common integral:

\int \frac{1}{- x ^{2} + 1} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

= \frac{1}{3} (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5})

Simplify

\frac{1}{3} (- [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{4}^{5}) : - \frac{1}{3} [\frac{ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{2}]_{4}^{5}

= - \frac{1}{3} [\frac{ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣}{2}]_{4}^{5}

Compute the boundaries:

\frac{ln ( 6 ) - 2 ln ( 2 ) - ln ( 5 ) + ln ( 3 )}{2}

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

Simplify

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

What is the integral from 4 to 5 of 1/(3x^2-3) ?
The integral from 4 to 5 of 1/(3x^2-3) is -(ln(3/2)-ln(5)+ln(3))/6