integral of (8t-3)/(t+1)

解答

\int \frac{8 t - 3}{t + 1} d t

8 (t + 1) - 11 ln ∣ t + 1 ∣ + C

求解步骤

\int \frac{8 t - 3}{t + 1} d t

使用换元积分法

= \int \frac{8 u - 11}{u} d u

乘开

\frac{8 u - 11}{u} : 8 - \frac{11}{u}

= \int 8 - \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

= \int 8 d u - \int \frac{11}{u} d u

\int 8 d u = 8 u

\int \frac{11}{u} d u = 11 ln ∣ u ∣

= 8 u - 11 ln ∣ u ∣

u = t + 1

代回

= 8 (t + 1) - 11 ln ∣ t + 1 ∣

解答补常数

= 8 (t + 1) - 11 ln ∣ t + 1 ∣ + C

x \to 4 lim (x^{2} + 5 - x)

t^{2} y^{^{''}} + 3 t y^{^{'}} - 8 y = 0

\frac{\partial}{\partial y} (x^{4} + x^{2} y)

\frac{d}{d x} (\frac{x ^{2} + 3 x}{6})

\int_{0}^{16} x - 4 d x