What is the integral from 0 to 1 of (12)/(x^2-4) ?

The integral from 0 to 1 of (12)/(x^2-4) is -3ln(3)

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integral from 0 to 1 of (12)/(x^2-4)

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

- 3 ln (3)

Decimal

- 3.29583 \dots

Solution steps

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{1} \frac{1}{x ^{2} - 4} d x

Apply Integral Substitution

= 12 \cdot \int_{0}^{\frac{1}{2}} \frac{1}{2 ( u ^{2} - 1 )} d u

Take the constant out:

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

= 12 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \frac{1}{u ^{2} - 1} d u

Factor

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

= 12 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 12 \cdot \frac{1}{2} (- \int_{0}^{\frac{1}{2}} \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

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

= 12 \cdot \frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}})

Simplify

12 \cdot \frac{1}{2} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{1}{2}}) : - 6 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{2}}

= - 6 [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{0}^{\frac{1}{2}}

Compute the boundaries:

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

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

Simplify

= - 3 ln (3)

What is the integral from 0 to 1 of (12)/(x^2-4) ?
The integral from 0 to 1 of (12)/(x^2-4) is -3ln(3)