integral of 1/(sqrt(6-x))

解答

\int \frac{1}{6 - x} d x

- 2 6 - x + C

求解步骤

\int \frac{1}{6 - x} d x

使用换元积分法

= \int - \frac{1}{u} d u

提出常数:

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

= - \int \frac{1}{u} d u

使用幂法则

= - 2 u^{\frac{1}{2}}

u = 6 - x

代回

= - 2 (6 - x)^{\frac{1}{2}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

= - 2 6 - x

解答补常数

= - 2 6 - x + C

\int 1 + x^{4} d x

\frac{d}{d x} (ln ∣ sec (x) ∣)

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

\frac{\partial}{\partial x} (2 (x^{2} + 4)^{- \frac{1}{2}})

l a pl a ce t r an s f or m 6 e^{- 3 t} - t^{2} + 2 t - 8