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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用换元积分法

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

提出常数:

\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

对

- u^{2} + 1 + u

配方:

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

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

使用换元积分法

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

提出常数:

\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

乘开

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

使用积分加法定则:

\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))

代回

= - 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))

化简

- 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)

解答补常数

= 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

作图

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