integral of (9t-5)/(t+1)

解答

\int \frac{9 t - 5}{t + 1} d t

9 (t + 1) - 14 ln ∣ t + 1 ∣ + C

求解步骤

\int \frac{9 t - 5}{t + 1} d t

使用换元积分法

= \int \frac{9 u - 14}{u} d u

乘开

\frac{9 u - 14}{u} : 9 - \frac{14}{u}

= \int 9 - \frac{14}{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 9 d u - \int \frac{14}{u} d u

\int 9 d u = 9 u

\int \frac{14}{u} d u = 14 ln ∣ u ∣

= 9 u - 14 ln ∣ u ∣

u = t + 1

代回

= 9 (t + 1) - 14 ln ∣ t + 1 ∣

解答补常数

= 9 (t + 1) - 14 ln ∣ t + 1 ∣ + C

x \to 0 lim (\frac{x - 32}{x - 4})

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

d er i v a t i v e f (x) = \frac{x ^{2} + 2 x}{x ^{2} - 1}

\int \frac{1}{1 + 14 x} d x

f (x) = 4 x + 1