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受欢迎的微积分 >

integral of 1/(25e^{-4x)+e^{4x}}

解答

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

解答

- \frac{1}{20} arctan (5 e^{- 4 x}) + C

求解步骤

\int \frac{1}{25 e ^{- 4 x} + e ^{4 x}} d x

使用换元积分法

= \int - \frac{1}{4 ( 25 u ^{2} + 1 )} d u

提出常数:

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

= - \frac{1}{4} \cdot \int \frac{1}{25 u ^{2} + 1} d u

使用换元积分法

= - \frac{1}{4} \cdot \int \frac{1}{5 ( 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}{4} \cdot \frac{1}{5} \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}{4} \cdot \frac{1}{5} arctan (v)

代回

= - \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x})

化简

- \frac{1}{4} \cdot \frac{1}{5} arctan (5 e^{- 4 x}) : - \frac{1}{20} arctan (5 e^{- 4 x})

= - \frac{1}{20} arctan (5 e^{- 4 x})

解答补常数

= - \frac{1}{20} arctan (5 e^{- 4 x}) + C

作图

流行的例子

derivative f(x)=(5x^6+2x^3)^4

d er i v a t i v e f (x) = (5 x^{6} + 2 x^{3})^{4}

integral of 1/(-x-1)

\int \frac{1}{- x - 1} d x

(\partial)/(\partial x)(2(x-y)e^{2y+5x^2})

\frac{\partial}{\partial x} (2 (x - y) e^{2 y + 5 x^{2}})

f^'(t)-f(t)=-7t,f(2)=-4

f^{^{'}} (t) - f (t) = - 7 t, f (2) = - 4

integral of (2y)/(x^2+y^2)

\int \frac{2 y}{x ^{2} + y ^{2}} d x