integral of (e^t+1)^2

解答

\int (e^{t} + 1)^{2} d t

\frac{1}{2} e^{2 t} + 2 e^{t} + t + C

求解步骤

\int (e^{t} + 1)^{2} d t

使用换元积分法

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

乘开

\frac{( u + 1 ) ^{2}}{u} : u + 2 + \frac{1}{u}

= \int u + 2 + \frac{1}{u} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u + \int 2 d u + \int \frac{1}{u} d u

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + ln ∣ u ∣

u = e^{t}

代回

= \frac{( e ^{t} ) ^{2}}{2} + 2 e^{t} + ln e^{t}

化简

\frac{( e ^{t} ) ^{2}}{2} + 2 e^{t} + ln e^{t} : \frac{1}{2} e^{2 t} + 2 e^{t} + t

= \frac{1}{2} e^{2 t} + 2 e^{t} + t

解答补常数

= \frac{1}{2} e^{2 t} + 2 e^{t} + t + C

x \to 2 lim (\frac{1}{x + 1})

in v erse l a pl a ce \frac{3 s}{s ^{2} + 3 s + 16}

\frac{d}{d x} (x^{3} tan (x))

\frac{\partial}{\partial x} (e^{x + y + 4})

\int_{- 1}^{1} x sin (π x) d x