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

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

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

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

- \frac{1}{2} (ln (3) - ln (2))

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= - \frac{1}{2} (ln (3) - ln (2))

\int_{4}^{t} \frac{12}{x} d x

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\int_{0}^{6} 6 x - x^{2} d x

\int_{3}^{4} ln^{2} (x) d x

\int_{0}^{1} (x^{3} + 1) d x

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