integral of x^22^{-x}

解答

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

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

求解步骤

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

使用换元积分法

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

提出常数:

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

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

使用分布积分法

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

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

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

u = - x

代回

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

化简

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

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

解答补常数

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

\int \frac{1 + x}{1 + 3 x} d x

\int \frac{10 t}{t ^{2} + 9} d t

\int 3 x^{3} + 1 \cdot 3 x^{2} d x

\int \frac{3 t + 1}{t ^{2} ( t ^{2} - 1 )} d t

\int \frac{2 x + 1}{x ^{2} - 2 x + 3} d x