integral of 1/(a^x)

解答

\int \frac{1}{a ^{x}} d x

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

求解步骤

\int \frac{1}{a ^{x}} d x

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{a ^{x}} = a^{- x}

= \int a^{- x} d x

使用换元积分法

= \int - a^{u} d u

提出常数:

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

= - \int a^{u} d u

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

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

u = - x

代回

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

解答补常数

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

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