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人気のある微分積分 >

integral from 0 to 1 of 5/(4x+1)

解

\int_{0}^{1} \frac{5}{4 x + 1} d x

解

\frac{5}{4} ln (5)

+1

十進法表記

2.01179 \dots

解答ステップ

\int_{0}^{1} \frac{5}{4 x + 1} d x

定数を除く:

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

= 5 \cdot \int_{0}^{1} \frac{1}{4 x + 1} d x

u置換積分法を適用する

= 5 \cdot \int_{1}^{5} \frac{1}{4 u} d u

定数を除く:

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

= 5 \cdot \frac{1}{4} \cdot \int_{1}^{5} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{5}

簡素化

5 \cdot \frac{1}{4} [ln ∣ u ∣]_{1}^{5} : \frac{5}{4} [ln ∣ u ∣]_{1}^{5}

= \frac{5}{4} [ln ∣ u ∣]_{1}^{5}

限度を計算する:

ln (5)

= \frac{5}{4} ln (5)

グラフ

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