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人気のある微分積分 >

integral of 5/(x(x^4+9))

解

\int \frac{5}{x ( x ^{4} + 9 )} d x

解

\frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

解答ステップ

\int \frac{5}{x ( x ^{4} + 9 )} d x

定数を除く:

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

= 5 \cdot \int \frac{1}{x ( x ^{4} + 9 )} d x

u置換積分法を適用する

= 5 \cdot \int \frac{1}{4 u ( u + 9 )} 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 \frac{1}{u ( u + 9 )} d u

以下の部分分数を得る:

\frac{1}{u ( u + 9 )} : \frac{1}{9 u} - \frac{1}{9 ( u + 9 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{9 u} - \frac{1}{9 ( u + 9 )} 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

= 5 \cdot \frac{1}{4} (\int \frac{1}{9 u} d u - \int \frac{1}{9 ( u + 9 )} d u)

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u + 9 )} d u = \frac{1}{9} ln ∣ u + 9 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u + 9 ∣)

代用を戻す

u = x^{4}

= 5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9)

簡素化

5 \cdot \frac{1}{4} (\frac{1}{9} ln x^{4} - \frac{1}{9} ln x^{4} + 9) : \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9

定数を解答に追加する

= \frac{5}{9} ln (x) - \frac{5}{36} ln x^{4} + 9 + C

グラフ

人気の例

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\frac{d}{d x} (8 x + 1)

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