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受欢迎的微积分 >

integral of (e^x)/(sqrt(1-25e^{2x))}

解答

\int \frac{e ^{x}}{1 - 25 e ^{2 x}} d x

解答

i \frac{1}{5} ln tan (arcsec (1 - 25 e^{2 x})) + sec (arcsec (1 - 25 e^{2 x})) + C

求解步骤

\int \frac{e ^{x}}{1 - 25 e ^{2 x}} d x

使用换元积分法

= \int i \frac{- 1 + u ^{2}}{- 5 + 5 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}}{- 5 + 5 u ^{2}} d u

化简

\frac{- 1 + u ^{2}}{- 5 + 5 u ^{2}} : \frac{1}{5 - 1 + u ^{2}}

= i \cdot \int \frac{1}{5 - 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}{5} \cdot \int \frac{1}{- 1 + u ^{2}} d u

使用三角换元法

= i \frac{1}{5} \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}{5} ln ∣ tan (v) + sec (v) ∣

代回

= i \frac{1}{5} ln tan (arcsec (1 - 25 e^{2 x})) + sec (arcsec (1 - 25 e^{2 x}))

解答补常数

= i \frac{1}{5} ln tan (arcsec (1 - 25 e^{2 x})) + sec (arcsec (1 - 25 e^{2 x})) + C

流行的例子

(\partial)/(\partial x)(x/(x^2+y^2))

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

(\partial)/(\partial x)(ln(x+5y+3z))

\frac{\partial}{\partial x} (ln (x + 5 y + 3 z))

(\partial)/(\partial x)(x^3+3xy^2-3x)

\frac{\partial}{\partial x} (x^{3} + 3 x y^{2} - 3 x)

integral from 2.5 to 0.5 of (23.75-7.5v)

\int_{2.5}^{0.5} (23.75 - 7.5 v) d v

(\partial)/(\partial y)(8x-(x+y)^3)

\frac{\partial}{\partial y} (8 x - (x + y)^{3})