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

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

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integral from 3 to 5 of 1/(x^2-3x+2)

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

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

Decimal

0.40546 \dots

Solution steps

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

Complete the square

x^{2} - 3 x + 2 : (x - \frac{3}{2})^{2} - \frac{1}{4}

= \int_{3}^{5} \frac{1}{( x - \frac{3}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

: \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{4}{4 u ^{2} - 1} d u

= \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{4}{4 u ^{2} - 1} 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{7}{2}} \frac{1}{4 u ^{2} - 1} d u

Factor

4 u^{2} - 1 : - (- 4 u^{2} + 1)

= 4 \cdot \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{1}{- ( - 4 u ^{2} + 1 )} d u

Take the constant out:

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

= 4 (- \int_{\frac{3}{2}}^{\frac{7}{2}} \frac{1}{- 4 u ^{2} + 1} d u)

Apply Integral Substitution

= 4 (- \int_{3}^{7} \frac{1}{2 ( - v ^{2} + 1 )} d v)

Take the constant out:

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

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

Simplify

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

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

Compute the boundaries:

\frac{2 ln ( 2 ) - ln ( 6 )}{2}

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

Simplify

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

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