integral of (1+e^{2x})/(e^{x-1)}

解答

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

e (- \frac{1}{e ^{x}} + e^{x}) + C

求解步骤

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

\frac{1 + e ^{2 x}}{e ^{x - 1}} = \frac{1}{e ^{- 1}} \frac{1 + e ^{2 x}}{e ^{x}}

= \int \frac{1}{e ^{- 1}} \cdot \frac{1 + e ^{2 x}}{e ^{x}} d x

提出常数:

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

= \frac{1}{e ^{- 1}} \cdot \int \frac{1 + e ^{2 x}}{e ^{x}} d x

\frac{1}{e ^{- 1}} = e

= e \cdot \int \frac{1 + e ^{2 x}}{e ^{x}} d x

\frac{1 + e ^{2 x}}{e ^{x}} = (\frac{1}{e ^{x}} + \frac{e ^{2 x}}{e ^{x}})

= e \cdot \int \frac{1}{e ^{x}} + \frac{e ^{2 x}}{e ^{x}} d x

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

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

\int \frac{1}{e ^{x}} d x = - \frac{1}{e ^{x}}

\int \frac{e ^{2 x}}{e ^{x}} d x = e^{x}

= e (- \frac{1}{e ^{x}} + e^{x})

解答补常数

= e (- \frac{1}{e ^{x}} + e^{x}) + C

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

\int x + \frac{1}{x} - 7 d x

x \to 7 lim (5 x^{2} - 5 x + 6)

x \to 0 lim (\frac{1}{1 + cos ( x )})

\frac{d}{d x} (\frac{3}{4} (x^{2} - 1)^{\frac{2}{3}})