integral of y/(y+3)

解答

\int \frac{y}{y + 3} d y

y + 3 - 3 ln ∣ y + 3 ∣ + C

求解步骤

\int \frac{y}{y + 3} d y

使用换元积分法

= \int \frac{u - 3}{u} d u

乘开

\frac{u - 3}{u} : 1 - \frac{3}{u}

= \int 1 - \frac{3}{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{3}{u} d u

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

= u - 3 ln ∣ u ∣

u = y + 3

代回

= y + 3 - 3 ln ∣ y + 3 ∣

解答补常数

= y + 3 - 3 ln ∣ y + 3 ∣ + C

x \to 0 lim (\frac{5 x ^{2}}{x ^{4}})

\int_{0}^{2} x^{3} 3 x + 1 d x

t a y l or ln (x), at 4

f (x) = \frac{- x ( x + 4 )}{( x ^{2} - x - 2 ) ^{2}}

n = 1 \sum \infty \frac{( e ^{n} )}{n}