integral of y/(y+1)

解答

\int \frac{y}{y + 1} d y

y + 1 - ln ∣ y + 1 ∣ + C

求解步骤

\int \frac{y}{y + 1} d y

使用换元积分法

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

乘开

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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

u = y + 1

代回

= y + 1 - ln ∣ y + 1 ∣

解答补常数

= y + 1 - ln ∣ y + 1 ∣ + C

x y (\frac{d y}{d x}) = 3

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