integral of 4^{-x}

解答

\int 4^{- x} d x

- \frac{2 ^{- 2 x - 1}}{ln ( 2 )} + C

求解步骤

\int 4^{- x} d x

使用换元积分法

= \int - 4^{u} d u

提出常数:

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

= - \int 4^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - \frac{4 ^{u}}{ln ( 4 )}

u = - x

代回

= - \frac{4 ^{- x}}{ln ( 4 )}

化简

- \frac{4 ^{- x}}{ln ( 4 )} : - \frac{2 ^{- 2 x - 1}}{ln ( 2 )}

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

解答补常数

= - \frac{2 ^{- 2 x - 1}}{ln ( 2 )} + C

\frac{\partial}{\partial x} (a x + b y + c)

\frac{\partial}{\partial x} (- 6 sin (2 x + y))

\int_{0}^{N} e^{- s t} sin (4 t) d t

x \to - 2 lim (- 2 x^{2} + 4 x - 9)

y^{^{'}} = \frac{2 x ^{3}}{y ^{2}}, y (2) = 1