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人気のある微分積分 >

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

解

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

解

ln 4 x + 4 - x + 1 - \frac{1}{5} (ln \frac{2 1 - x - 1}{5} + 1 - ln \frac{2 1 - x - 1}{5} - 1) + C

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

平方完成する

- u^{2} + 1 + u : - (u - \frac{1}{2})^{2} + \frac{5}{4}

= - 2 \cdot \int \frac{u}{- ( u - \frac{1}{2} ) ^{2} + \frac{5}{4}} d u

u置換積分法を適用する

= - 2 \cdot \int \frac{2 ( 2 v + 1 )}{- 4 v ^{2} + 5} d v

定数を除く:

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

= - 2 \cdot 2 \cdot \int \frac{2 v + 1}{- 4 v ^{2} + 5} d v

拡張

\frac{2 v + 1}{- 4 v ^{2} + 5} : \frac{2 v}{- 4 v ^{2} + 5} + \frac{1}{- 4 v ^{2} + 5}

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 2 (\int \frac{2 v}{- 4 v ^{2} + 5} d v + \int \frac{1}{- 4 v ^{2} + 5} d v)

\int \frac{2 v}{- 4 v ^{2} + 5} d v = - \frac{1}{4} ln - 4 v^{2} + 5

\int \frac{1}{- 4 v ^{2} + 5} d v = \frac{1}{4 5} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1)

= - 2 \cdot 2 (- \frac{1}{4} ln - 4 v^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} v + 1 - ln \frac{2}{5} v - 1))

代用を戻す

= - 2 \cdot 2 (- \frac{1}{4} ln - 4 (1 - x - \frac{1}{2})^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} (1 - x - \frac{1}{2}) + 1 - ln \frac{2}{5} (1 - x - \frac{1}{2}) - 1))

簡素化

- 2 \cdot 2 (- \frac{1}{4} ln - 4 (1 - x - \frac{1}{2})^{2} + 5 + \frac{1}{4 5} (ln \frac{2}{5} (1 - x - \frac{1}{2}) + 1 - ln \frac{2}{5} (1 - x - \frac{1}{2}) - 1)) : ln 4 x + 4 - x + 1 - \frac{1}{5} (ln \frac{2 1 - x - 1}{5} + 1 - ln \frac{2 1 - x - 1}{5} - 1)

= ln 4 x + 4 - x + 1 - \frac{1}{5} (ln \frac{2 1 - x - 1}{5} + 1 - ln \frac{2 1 - x - 1}{5} - 1)

定数を解答に追加する

= ln 4 x + 4 - x + 1 - \frac{1}{5} (ln \frac{2 1 - x - 1}{5} + 1 - ln \frac{2 1 - x - 1}{5} - 1) + C

グラフ

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