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人気のある微分積分 >

integral of (15)/(5x+xsqrt(x))

解

\int \frac{15}{5 x + x x} d x

解

30 (\frac{1}{5} ln x - \frac{1}{5} ln x + 5) + C

解答ステップ

\int \frac{15}{5 x + x x} d x

定数を除く:

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

= 15 \cdot \int \frac{1}{5 x + x x} d x

u置換積分法を適用する

= 15 \cdot \int \frac{2}{u ( u + 5 )} d u

定数を除く:

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

= 15 \cdot 2 \cdot \int \frac{1}{u ( u + 5 )} d u

以下の部分分数を得る:

\frac{1}{u ( u + 5 )} : \frac{1}{5 u} - \frac{1}{5 ( u + 5 )}

= 15 \cdot 2 \cdot \int \frac{1}{5 u} - \frac{1}{5 ( u + 5 )} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 15 \cdot 2 (\int \frac{1}{5 u} d u - \int \frac{1}{5 ( u + 5 )} d u)

\int \frac{1}{5 u} d u = \frac{1}{5} ln ∣ u ∣

\int \frac{1}{5 ( u + 5 )} d u = \frac{1}{5} ln ∣ u + 5 ∣

= 15 \cdot 2 (\frac{1}{5} ln ∣ u ∣ - \frac{1}{5} ln ∣ u + 5 ∣)

代用を戻す

u = x

= 15 \cdot 2 (\frac{1}{5} ln x - \frac{1}{5} ln x + 5)

簡素化

= 30 (\frac{1}{5} ln x - \frac{1}{5} ln x + 5)

定数を解答に追加する

= 30 (\frac{1}{5} ln x - \frac{1}{5} ln x + 5) + C

グラフ

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