integral of sech(1/4)xtanh(1/4)x

解答

\int sech (\frac{1}{4}) x tanh (\frac{1}{4}) x d x

\frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 4 e 4 e + 1} \cdot \frac{x ^{3}}{3} + C

求解步骤

\int sech (\frac{1}{4}) x tanh (\frac{1}{4}) x d x

提出常数:

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

= sech (\frac{1}{4}) tanh (\frac{1}{4}) \cdot \int xx d x

sech (\frac{1}{4}) tanh (\frac{1}{4}) = \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 e ^{\frac{1}{4}} 4 e + 1}

= \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 e ^{\frac{1}{4}} 4 e + 1} \cdot \int xx d x

化简

xx : x^{2}

= \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 e ^{\frac{1}{4}} 4 e + 1} \cdot \int x^{2} d x

使用幂法则

= \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 e ^{\frac{1}{4}} 4 e + 1} \cdot \frac{x ^{3}}{3}

使用根式运算法则:

a^{\frac{1}{n}} = n a

= \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 4 e 4 e + 1} \cdot \frac{x ^{3}}{3}

解答补常数

= \frac{2 e ^{\frac{3}{4}} - 2 4 e}{e + 2 4 e 4 e + 1} \cdot \frac{x ^{3}}{3} + C

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