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

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

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integral from 0 to 1 of 3/(x^2-x-2)

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

- 2 ln (2)

Decimal Notation

- 1.38629 \dots

Solution steps

\int_{0}^{1} \frac{3}{x ^{2} - x - 2} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{1}{x ^{2} - x - 2} d x

Complete the square

x^{2} - x - 2 : (x - \frac{1}{2})^{2} - \frac{9}{4}

= 3 \cdot \int_{0}^{1} \frac{1}{( x - \frac{1}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= 3 \cdot 4 \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \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

= 3 \cdot 4 \cdot \frac{1}{6} \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \frac{1}{v ^{2} - 1} d v

Factor

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

= 3 \cdot 4 \cdot \frac{1}{6} \cdot \int_{- \frac{1}{3}}^{\frac{1}{3}} \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

= 3 \cdot 4 \cdot \frac{1}{6} (- \int_{- \frac{1}{3}}^{\frac{1}{3}} \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}

= 3 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{1}{3}})

Simplify

3 \cdot 4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{- \frac{1}{3}}^{\frac{1}{3}}) : - 2 [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{- \frac{1}{3}}^{\frac{1}{3}}

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

Compute the boundaries:

\frac{- 2 ln ( \frac{2}{3} ) + 2 ln ( \frac{4}{3} )}{2}

= - 2 \cdot \frac{- 2 ln ( \frac{2}{3} ) + 2 ln ( \frac{4}{3} )}{2}

Simplify

= - 2 ln (2)

\frac{3}{x ^{2}}

n = 1 \sum \infty (- 1)^{n} e^{- n}

x \to 2 lim (3 x - 9)

\frac{d}{d x} (- 125 cos (5 x))

\int_{\frac{1}{2}}^{\frac{3}{4}} \frac{1}{u ^{2}} d u

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