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受欢迎的微积分 >

integral of ((e^x)/(sqrt(1-9e^{2x))})

解答

\int (\frac{e ^{x}}{1 - 9 e ^{2 x}}) d x

解答

i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x})) + C

求解步骤

\int \frac{e ^{x}}{1 - 9 e ^{2 x}} d x

使用换元积分法

= \int i \frac{- 1 + u ^{2}}{- 3 + 3 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}}{- 3 + 3 u ^{2}} d u

化简

\frac{- 1 + u ^{2}}{- 3 + 3 u ^{2}} : \frac{1}{3 - 1 + u ^{2}}

= i \cdot \int \frac{1}{3 - 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}{3} \cdot \int \frac{1}{- 1 + u ^{2}} d u

使用三角换元法

= i \frac{1}{3} \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}{3} ln ∣ tan (v) + sec (v) ∣

代回

= i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x}))

解答补常数

= i \frac{1}{3} ln tan (arcsec (1 - 9 e^{2 x})) + sec (arcsec (1 - 9 e^{2 x})) + C

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\int x^{2} \cdot 4 2 x^{3} + 9 d x

integral of (4x^3-3x^2+4x)/(x^2+1)

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integral of (5e^{x^{-2}})/(x^3)

\int \frac{5 e ^{x^{- 2}}}{x ^{3}} d x

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\int e^{x} + 6 d x

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\int 4 e^{4 z}