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人気のある微分積分 >

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

解

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

解

\frac{1}{2} (ln (5) - ln (2))

+1

十進法表記

0.45814 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{2}^{5} \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}^{5} \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}^{5}

限度を計算する:

ln (5) - ln (2)

= \frac{1}{2} (ln (5) - ln (2))

グラフ

人気の例

integral from 1 to 8 of (3(ln(x))^2)/x

\int_{1}^{8} \frac{3 ( ln ( x ) ) ^{2}}{x} d x

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