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人気のある微分積分 >

integral of (3sqrt(x))/(x-1)

解

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

解

- 6 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x) + C

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

因数

u^{2} - 1 : - (- u^{2} + 1)

= 3 \cdot 2 \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

定数を除く:

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

= 3 \cdot 2 (- \int \frac{u ^{2}}{- u ^{2} + 1} d u)

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= 3 \cdot 2 (- \int \frac{1}{- u ^{2} + 1} - 1 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

= 3 \cdot 2 (- (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

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

\int 1 d u = u

= 3 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

代用を戻す

u = x

= 3 \cdot 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x))

簡素化

3 \cdot 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x)) : - 6 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x)

= - 6 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x)

定数を解答に追加する

= - 6 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x) + C

グラフ

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