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

The integral from 2 to 3 of 2/(t^2-1) is -ln(2)+ln(3)

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integral from 2 to 3 of 2/(t^2-1)

\int_{2}^{3} \frac{2}{t ^{2} - 1} d t

- ln (2) + ln (3)

Decimal

0.40546 \dots

Solution steps

\int_{2}^{3} \frac{2}{t ^{2} - 1} d t

Take the constant out:

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

= 2 \cdot \int_{2}^{3} \frac{1}{t ^{2} - 1} d t

Factor

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

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

Take the constant out:

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

= 2 (- \int_{2}^{3} \frac{1}{- t ^{2} + 1} d t)

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - ln (2) + ln (3)

What is the integral from 2 to 3 of 2/(t^2-1) ?
The integral from 2 to 3 of 2/(t^2-1) is -ln(2)+ln(3)