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

解答

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

- 1 - x + 2 ln ∣ 1 + x ∣ + C

求解步骤

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

使用换元积分法

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

乘开

\frac{- u + 2}{u} : - 1 + \frac{2}{u}

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

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

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

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= - u + 2 ln ∣ u ∣

u = 1 + x

代回

= - (1 + x) + 2 ln ∣ 1 + x ∣

化简

= - 1 - x + 2 ln ∣ 1 + x ∣

解答补常数

= - 1 - x + 2 ln ∣ 1 + x ∣ + C

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