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受欢迎的微积分 >

integral of e^{tx-x}

解答

\int e^{t x - x} d x

解答

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

求解步骤

\int e^{t x - x} d x

乘以

- x + t x

的共轭根式

: \frac{t ^{2} x ^{2} - x ^{2}}{t x + x}

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

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

= \int e^{x (t - 1)} d x

使用换元积分法

= \int \frac{e ^{u}}{t - 1} d u

提出常数:

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

= \frac{1}{t - 1} \cdot \int e^{u} d u

使用积分常见定则:

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

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

u = x (t - 1)

代回

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

解答补常数

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

作图

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