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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

- 0.34657 \dots

解答ステップ

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

u置換積分法を適用する

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

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

= - \int_{1}^{2} \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_{1}^{2} \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 ∣]_{1}^{2}

限度を計算する:

ln (2)

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

グラフ

人気の例

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

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

integral from 0 to 22 of pi(sqrt(x))^2

\int_{0}^{22} π (x)^{2} d x

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

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

integral from 1 to-1 of (x^2-1)

\int_{1}^{- 1} (x^{2} - 1) d x

integral from-2 to 2 of (x+2)2^x

\int_{- 2}^{2} (x + 2) 2^{x} d x