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人気のある微分積分 >

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

解

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

解

\frac{ln ( 3 )}{ln ( 5 )}

+1

十進法表記

0.68260 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{2}^{6} \frac{1}{ln ( 5 ) u} d u

定数を除く:

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

= \frac{1}{ln ( 5 )} \cdot \int_{2}^{6} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 5 )} [ln ∣ u ∣]_{2}^{6}

限度を計算する:

ln (6) - ln (2)

= \frac{1}{ln ( 5 )} (ln (6) - ln (2))

簡素化

= \frac{ln ( 3 )}{ln ( 5 )}

グラフ

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