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人気のある微分積分 >

integral of (2x^3)/(x^2-1)

解

\int \frac{2 x ^{3}}{x ^{2} - 1} d x

解

x^{2} - 1 + ln x^{2} - 1 + C

解答ステップ

\int \frac{2 x ^{3}}{x ^{2} - 1} d x

定数を除く:

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

= 2 \cdot \int \frac{x ^{3}}{x ^{2} - 1} d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\frac{u + 1}{u} : 1 + \frac{1}{u}

= 2 \cdot \frac{1}{2} \cdot \int 1 + \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{2} (u + ln ∣ u ∣)

代用を戻す

u = x^{2} - 1

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

簡素化

2 \cdot \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) : x^{2} - 1 + ln x^{2} - 1

= x^{2} - 1 + ln x^{2} - 1

定数を解答に追加する

= x^{2} - 1 + ln x^{2} - 1 + C

グラフ

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