integral of x^24^{-x}

解答

\int x^{2} 4^{- x} d x

- \frac{2 ^{- 2 x - 1} x ^{2}}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{- 2 x - 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{- 2 x - 2} + C

求解步骤

\int x^{2} \cdot 4^{- x} d x

使用换元积分法

= \int - 4^{u} u^{2} d u

提出常数:

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

= - \int 4^{u} u^{2} d u

使用分布积分法

= - (\frac{2 ^{2 u - 1} u ^{2}}{ln ( 2 )} - \int \frac{4 ^{u} u}{ln ( 2 )} d u)

\int \frac{4 ^{u} u}{ln ( 2 )} d u = \frac{1}{ln ( 2 )} (\frac{2 ^{2 u - 1} u}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2 u - 2})

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

u = - x

代回

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

化简

- (\frac{2 ^{2 (- x) - 1} ( - x ) ^{2}}{ln ( 2 )} - \frac{1}{ln ( 2 )} (\frac{2 ^{2 (- x) - 1} ( - x )}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2 (- x) - 2})) : - \frac{2 ^{- 2 x - 1} x ^{2}}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{- 2 x - 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{- 2 x - 2}

= - \frac{2 ^{- 2 x - 1} x ^{2}}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{- 2 x - 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{- 2 x - 2}

解答补常数

= - \frac{2 ^{- 2 x - 1} x ^{2}}{ln ( 2 )} - \frac{1}{ln ^{2} ( 2 )} \cdot 2^{- 2 x - 1} x - \frac{1}{ln ^{3} ( 2 )} \cdot 2^{- 2 x - 2} + C

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

\int a^{n y} d y

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\int \frac{x ^{2} + x + 8}{x ^{2} + 3 x - 4} d x

\int \frac{12 x ^{8} - x}{x ^{3}} d x