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

The integral from 2 to 3 of 1/(x^2+x-2) is -(ln(5/8))/3

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

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

- \frac{ln ( \frac{5}{8} )}{3}

Decimal

0.15666 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{\frac{5}{2}}^{\frac{7}{2}} \frac{4}{4 u ^{2} - 9} d u

= \int_{\frac{5}{2}}^{\frac{7}{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{5}{2}}^{\frac{7}{2}} \frac{1}{4 u ^{2} - 9} d u

Apply Integral Substitution

= 4 \cdot \int_{\frac{5}{3}}^{\frac{7}{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_{\frac{5}{3}}^{\frac{7}{3}} \frac{1}{v ^{2} - 1} d v

Factor

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

= 4 \cdot \frac{1}{6} \cdot \int_{\frac{5}{3}}^{\frac{7}{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_{\frac{5}{3}}^{\frac{7}{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}]_{\frac{5}{3}}^{\frac{7}{3}})

Simplify

4 \cdot \frac{1}{6} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{\frac{7}{3}}) : - \frac{2}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{\frac{7}{3}}

= - \frac{2}{3} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{\frac{5}{3}}^{\frac{7}{3}}

Compute the boundaries:

\frac{ln ( \frac{10}{3} ) - ln ( \frac{4}{3} ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

= - \frac{2}{3} \cdot \frac{ln ( \frac{10}{3} ) - ln ( \frac{4}{3} ) - ln ( \frac{8}{3} ) + ln ( \frac{2}{3} )}{2}

Simplify

= - \frac{ln ( \frac{5}{8} )}{3}

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What is the integral from 2 to 3 of 1/(x^2+x-2) ?
The integral from 2 to 3 of 1/(x^2+x-2) is -(ln(5/8))/3