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

The integral from-1 to 1 of 1/(x^2-x-2) is diverges

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

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

diverges

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

= \int_{- \frac{3}{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

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

Apply Integral Substitution

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

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

Factor

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

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

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

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

Simplify

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

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

Compute the boundaries:

diverges

= diverges

\int ln (1 + x^{4}) d x

n \to \infty lim (- 1)

\int_{1}^{2} - \frac{1}{16 z} + \frac{1}{4 z} d z

20 x (2 x^{2} + 7)^{4}

\frac{d}{d x} (- 5 x^{2} + 6 x)

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