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

解答

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

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

求解步骤

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

使用换元积分法

= \int \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

使用换元积分法

= \int - e^{v} d v

提出常数:

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

= - \int e^{v} d v

使用积分常见定则:

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

= - e^{v}

代回

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

解答补常数

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

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