integral of 1/(2e^u)

解答

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

- \frac{1}{2 e ^{u}} + C

求解步骤

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

提出常数:

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

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

使用换元积分法

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

使用幂法则

= \frac{1}{2} (- \frac{1}{v})

v = e^{u}

代回

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

化简

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

= - \frac{1}{2 e ^{u}}

解答补常数

= - \frac{1}{2 e ^{u}} + C

\int \frac{1}{2} x sinh (2 x) d x

\int (4 x^{2} - 5 x + 1) d x

\int (2 x^{2} - 3 x + 1) e^{- 3 x} d x

\int (x^{2} - 6 x + 10) d x

\int sin^{2} (10 t) sec^{4} (10 t) d t