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人気のある微分積分 >

integral of 1/(x(2+x^{10))}

解

\int \frac{1}{x ( 2 + x ^{10} )} d x

解

\frac{1}{10} (5 ln (x) - \frac{1}{2} ln x^{10} + 2) + C

解答ステップ

\int \frac{1}{x ( 2 + x ^{10} )} d x

u置換積分法を適用する

= \int \frac{1}{10 u ( 2 + u )} d u

定数を除く:

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

= \frac{1}{10} \cdot \int \frac{1}{u ( 2 + u )} d u

以下の部分分数を得る:

\frac{1}{u ( 2 + u )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

= \frac{1}{10} \cdot \int \frac{1}{2 u} - \frac{1}{2 ( u + 2 )} 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

= \frac{1}{10} (\int \frac{1}{2 u} d u - \int \frac{1}{2 ( u + 2 )} d u)

\int \frac{1}{2 u} d u = \frac{1}{2} ln ∣ u ∣

\int \frac{1}{2 ( u + 2 )} d u = \frac{1}{2} ln ∣ u + 2 ∣

= \frac{1}{10} (\frac{1}{2} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 2 ∣)

代用を戻す

u = x^{10}

= \frac{1}{10} (\frac{1}{2} ln x^{10} - \frac{1}{2} ln x^{10} + 2)

\frac{1}{2} ln x^{10} = 5 ln (x)

= \frac{1}{10} (5 ln (x) - \frac{1}{2} ln x^{10} + 2)

定数を解答に追加する

= \frac{1}{10} (5 ln (x) - \frac{1}{2} ln x^{10} + 2) + C

グラフ

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