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人気のある微分積分 >

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

解

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

解

\frac{5 π}{2}

+1

十進法表記

7.85398 \dots

解答ステップ

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

u置換積分法を適用する

= 5 \cdot \int_{0}^{1} \frac{2}{1 + u ^{2}} d u

定数を除く:

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

= 5 \cdot 2 \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

共通積分を使用する:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 5 \cdot 2 [arctan (u)]_{0}^{1}

簡素化

= 10 [arctan (u)]_{0}^{1}

限度を計算する:

\frac{π}{4}

= 10 \cdot \frac{π}{4}

簡素化

= \frac{5 π}{2}

グラフ

人気の例

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