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人気のある微分積分 >

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

解

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

解

\frac{5}{4} (ln (x) - \frac{1}{4} ln x^{4} + 4) + C

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

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

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

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

代用を戻す

u = x^{4}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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