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

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

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

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

diverges

Solution steps

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

Apply u-substitution

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

Apply Integral Substitution

= 4 (- \int_{- 3}^{- 1} \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}^{- 1} \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}^{- 1})

Simplify

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

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

Compute the boundaries:

diverges

= diverges

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