integral of ((1-3x))/((3+2x))

解答

\int \frac{( 1 - 3 x )}{( 3 + 2 x )} d x

\frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣) + C

求解步骤

\int \frac{1 - 3 x}{3 + 2 x} d x

使用换元积分法

= \int \frac{- 3 u + 11}{4 u} d u

提出常数:

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

= \frac{1}{4} \cdot \int \frac{- 3 u + 11}{u} d u

乘开

\frac{- 3 u + 11}{u} : - 3 + \frac{11}{u}

= \frac{1}{4} \cdot \int - 3 + \frac{11}{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

= \frac{1}{4} (- \int 3 d u + \int \frac{11}{u} d u)

\int 3 d u = 3 u

\int \frac{11}{u} d u = 11 ln ∣ u ∣

= \frac{1}{4} (- 3 u + 11 ln ∣ u ∣)

u = 3 + 2 x

代回

= \frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣)

解答补常数

= \frac{1}{4} (- 3 (3 + 2 x) + 11 ln ∣ 3 + 2 x ∣) + C

\frac{y ^{^{'}} + y ^{2}}{x ^{2}} = 0

\frac{d}{d M} (M^{2} (\frac{C}{2} - \frac{M}{3}))

x \to 0 lim (e^{1 - \frac{1}{x}})

x \to 1 - lim (\frac{x + 2}{x - 1})

d er i v a t i v e 6 cos (θ) + 3 sin^{2} (θ)