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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{ln ( 3 )}{4}

Decimal

0.27465 \dots

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Apply Integral Substitution

= - \int_{- \frac{1}{2}}^{0} \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

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

Factor

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

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

= - \frac{1}{2} (- \int_{- \frac{1}{2}}^{0} \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}

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= \frac{ln ( 3 )}{4}

\int_{0}^{π} 1 sin^{2} (x) d x

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

\int_{3}^{9} \frac{1}{x ( ln ( x ) ) ^{8}} d x

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

\int_{- 2}^{2} 16 - x^{4} d x

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