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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用换元积分法

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

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

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

使用积分常见定则:

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

= \frac{1}{e ^{2}} e arctan (v)

代回

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

化简

\frac{1}{e ^{2}} e arctan (\frac{1}{e} e^{x}) : \frac{1}{e} arctan (e^{x - 1})

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

解答补常数

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

作图

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