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受欢迎的微积分 >

integral of (e^{2t})/(1+6e^{2t)+9e^{4t}}

解答

\int \frac{e ^{2 t}}{1 + 6 e ^{2 t} + 9 e ^{4 t}} d t

解答

- \frac{1}{6 ( 3 e ^{2 t} + 1 )} + C

求解步骤

\int \frac{e ^{2 t}}{1 + 6 e ^{2 t} + 9 e ^{4 t}} d t

使用换元积分法

= \int \frac{1}{2 ( 1 + 6 u + 9 u ^{2} )} 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}{1 + 6 u + 9 u ^{2}} d u

对

1 + 6 u + 9 u^{2}

配方:

9 (u + \frac{1}{3})^{2}

= \frac{1}{2} \cdot \int \frac{1}{9 ( u + \frac{1}{3} ) ^{2}} d u

使用换元积分法

= \frac{1}{2} \cdot \int \frac{1}{9 v ^{2}} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{9} \cdot \int \frac{1}{v ^{2}} d v

使用幂法则

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{v})

代回

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 t} + \frac{1}{3}})

化简

\frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 t} + \frac{1}{3}}) : - \frac{1}{6 ( 3 e ^{2 t} + 1 )}

= - \frac{1}{6 ( 3 e ^{2 t} + 1 )}

解答补常数

= - \frac{1}{6 ( 3 e ^{2 t} + 1 )} + C

作图

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