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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

- 0.20273 \dots

解答ステップ

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

u置換積分法を適用する

= \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

定数を除く:

\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

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} [ln ∣ u ∣]_{2}^{3}

限度を計算する:

ln (3) - ln (2)

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

グラフ

人気の例

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

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\int_{0}^{1} 2^{2 x} d x