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受欢迎的微积分 >

integral of (e^x)/(sqrt(1-16e^{2x))}

解答

\int \frac{e ^{x}}{1 - 16 e ^{2 x}} d x

解答

i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x})) + C

求解步骤

\int \frac{e ^{x}}{1 - 16 e ^{2 x}} d x

使用换元积分法

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

提出常数:

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

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

化简

\frac{- 1 + u ^{2}}{- 4 + 4 u ^{2}} : \frac{1}{4 - 1 + u ^{2}}

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

提出常数:

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

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

使用三角换元法

= i \frac{1}{4} \cdot \int sec (v) d v

使用积分常见定则:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= i \frac{1}{4} ln ∣ tan (v) + sec (v) ∣

代回

= i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x}))

解答补常数

= i \frac{1}{4} ln tan (arcsec (1 - 16 e^{2 x})) + sec (arcsec (1 - 16 e^{2 x})) + C

流行的例子

integral from 9 to 10 of 1/((x-9)^{4/3)}

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{4}{3}}} d x

integral from 0 to 6.25 of-x^2+5x

\int_{0}^{6.25} - x^{2} + 5 x d x

integral of 15x^{1/4}

\int 15 x^{\frac{1}{4}} d x

面积 \sqrt[3]{x}, 1/x ,x=27

a re a 3 x, \frac{1}{x}, x = 27

(\partial)/(\partial y)(y^{2x})

\frac{\partial}{\partial y} (y^{2 x})