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

解答

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

3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

求解步骤

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

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 = t + 1

代回

= 3 (t + 1) - 5 ln ∣ t + 1 ∣

解答补常数

= 3 (t + 1) - 5 ln ∣ t + 1 ∣ + C

d er i v a t i v e \frac{3}{x ^{5}}

y^{^{'}} + 5 y = 15

\frac{\partial}{\partial x} (- x - sin (y^{2}))

\int x (cos (x) - sin (x)) d x

\frac{d}{d x} (- \frac{( 2 x ^{2} + 1 )}{( 1 - x ^{2} ) ^{\frac{5}{2}}})