integral of e^{2t+e^t}

解答

\int e^{2 t + e^{t}} d t

e^{e^{t} + t} - e^{e^{t}} + C

求解步骤

\int e^{2 t + e^{t}} d t

使用换元积分法

= \int e^{u} u d u

使用分布积分法

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

\int e^{u} d u = e^{u}

= e^{u} u - e^{u}

u = e^{t}

代回

= e^{e^{t}} e^{t} - e^{e^{t}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

e^{e^{t}} e^{t} = e^{e^{t} + t}

= e^{e^{t} + t} - e^{e^{t}}

解答补常数

= e^{e^{t} + t} - e^{e^{t}} + C

\frac{d}{d x} ((5 x^{2} - 2 x + 1)^{- 3})

\int - \frac{5}{3 x} d x

d er i v a t i v e \frac{( 2 x ^{4} + 5 x ^{3} - 2 x - 8 )}{( x + 3 )}

\frac{\partial}{\partial x} (\frac{4 x - 3 t}{s + 2 x})

\frac{\partial}{\partial x} (x - cos (x))