integral of (y-3)/(y-2)

解答

\int \frac{y - 3}{y - 2} d y

- ln ∣ y - 2 ∣ + y + C

求解步骤

\int \frac{y - 3}{y - 2} d y

\frac{y - 3}{y - 2} = - \frac{1}{y - 2} + 1

= \int - \frac{1}{y - 2} + 1 d y

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 \frac{1}{y - 2} d y + \int 1 d y

\int \frac{1}{y - 2} d y = ln ∣ y - 2 ∣

\int 1 d y = y

= - ln ∣ y - 2 ∣ + y

解答补常数

= - ln ∣ y - 2 ∣ + y + C

y^{^{'}} = (x + y)^{2} - 1

tan (f (x)) (x) = - 4 x + \frac{1}{2} x - 5, at x = 1

\frac{\partial}{\partial x} (3 x^{2 y})

d er i v a t i v e s (t) = - 9.8 t^{2} + 2 t + 1

d er i v a t i v e y = (2 x + 3)^{5} + 10 x (2 x + 3)^{4}