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受欢迎的微积分 >

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

解答

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

解答

ln 4 x + 4 x - 1 - \frac{2}{3} arctan (\frac{2 x - 1 + 1}{3}) + C

求解步骤

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

使用换元积分法

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

使用换元积分法

= \int \frac{2 v}{v ^{2} + 1 + v} d v

提出常数:

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

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

对

v^{2} + 1 + v

配方:

(v + \frac{1}{2})^{2} + \frac{3}{4}

= 2 \cdot \int \frac{v}{( v + \frac{1}{2} ) ^{2} + \frac{3}{4}} d v

使用换元积分法

= 2 \cdot \int \frac{2 ( 2 w - 1 )}{4 w ^{2} + 3} d w

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 w - 1}{4 w ^{2} + 3} d w

乘开

\frac{2 w - 1}{4 w ^{2} + 3} : \frac{2 w}{4 w ^{2} + 3} - \frac{1}{4 w ^{2} + 3}

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 w}{4 w ^{2} + 3} d w - \int \frac{1}{4 w ^{2} + 3} d w)

\int \frac{2 w}{4 w ^{2} + 3} d w = \frac{1}{4} ln 4 w^{2} + 3

\int \frac{1}{4 w ^{2} + 3} d w = \frac{1}{2 3} arctan (\frac{2}{3} w)

= 2 \cdot 2 (\frac{1}{4} ln 4 w^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} w))

代回

= 2 \cdot 2 (\frac{1}{4} ln 4 (x - 1 + \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (x - 1 + \frac{1}{2})))

化简

2 \cdot 2 (\frac{1}{4} ln 4 (x - 1 + \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (x - 1 + \frac{1}{2}))) : ln 4 x + 4 x - 1 - \frac{2}{3} arctan (\frac{2 x - 1 + 1}{3})

= ln 4 x + 4 x - 1 - \frac{2}{3} arctan (\frac{2 x - 1 + 1}{3})

解答补常数

= ln 4 x + 4 x - 1 - \frac{2}{3} arctan (\frac{2 x - 1 + 1}{3}) + C

作图

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