integral of (3x+4)/(x+3)

解答

\int \frac{3 x + 4}{x + 3} d x

3 (x + 3) - 5 ln ∣ x + 3 ∣ + C

求解步骤

\int \frac{3 x + 4}{x + 3} d x

使用换元积分法

= \int \frac{3 u - 5}{u} d u

乘开

\frac{3 u - 5}{u} : 3 - \frac{5}{u}

= \int 3 - \frac{5}{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 3 d u - \int \frac{5}{u} d u

\int 3 d u = 3 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= 3 u - 5 ln ∣ u ∣

u = x + 3

代回

= 3 (x + 3) - 5 ln ∣ x + 3 ∣

解答补常数

= 3 (x + 3) - 5 ln ∣ x + 3 ∣ + C

\int_{0}^{1} - 10 (x^{2} + 1) e^{- x} d x

\frac{d y}{d x} = \frac{( x ^{2} - y )}{x}

\int x + 2 x d x d x

x \to 0 + lim (x^{2} - x - ln (x))

\int \frac{x ^{3} - 2 x - 2}{x ^{4} + 2 x ^{3} + 3 x ^{2}} d x