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受欢迎的微积分 >

integral of (sqrt(16-x^2))/x

解答

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

解答

- 2 ln \frac{1}{4} - x^{2} + 16 + 1 + 2 ln \frac{1}{4} - x^{2} + 16 - 1 + 16 - x^{2} + C

求解步骤

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

使用换元积分法

= \int - \frac{u ^{2}}{- u ^{2} + 16} d u

提出常数:

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

= - \int \frac{u ^{2}}{- u ^{2} + 16} d u

使用换元积分法

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

提出常数:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= - 4 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

代回

= - 4 \frac{ln \frac{16 - x ^{2}}{4} + 1}{2} - \frac{ln \frac{16 - x ^{2}}{4} - 1}{2} - \frac{16 - x ^{2}}{4}

化简

- 4 \frac{ln \frac{16 - x ^{2}}{4} + 1}{2} - \frac{ln \frac{16 - x ^{2}}{4} - 1}{2} - \frac{16 - x ^{2}}{4} : - 2 ln \frac{1}{4} - x^{2} + 16 + 1 + 2 ln \frac{1}{4} - x^{2} + 16 - 1 + 16 - x^{2}

= - 2 ln \frac{1}{4} - x^{2} + 16 + 1 + 2 ln \frac{1}{4} - x^{2} + 16 - 1 + 16 - x^{2}

解答补常数

= - 2 ln \frac{1}{4} - x^{2} + 16 + 1 + 2 ln \frac{1}{4} - x^{2} + 16 - 1 + 16 - x^{2} + C

作图

流行的例子

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\int_{- 1}^{0} 8 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

y^'=-(xy)/(x^2+y^2)

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

integral of (x^2)/(1+x^2)

\int \frac{x ^{2}}{1 + x ^{2}} d x

xy^'+4y=8x^4,y(1)=2

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

tangent x^2+3x+9,\at x=4

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